ELECTRICAL  ENGINEERING 

ADVANCED  COURSE 


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ELECTRICAL 
ENGINEERING 

ADVANCED  COURSE 


BY 

ERNST  JULIUS  BERG  Sc.  D. 

» i 

PROFESSOR  OP  ELECTRICAL  ENGINEERING  UNION 

COLLEGE,  SCHENECTADY,  N.  Y. 
AUTHOR  OF  "ELECTRICAL  ENGINEERING,"  FIRST  COURSE 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET.     NEW  YORK 


LONDON:  HILL  PUBLISHING  CO.,  LTD. 

6   &    8  BOUVEBIE   ST.,   E.    C. 

1916 


13 


COPYRIGHT,  1916,  BY  THE 
McGKAW-HiLL  BOOK  COMPANY,  INC. 


TUB     MAIT-IC     FKBSS     T  O  H  K 


PREFACE 

This  volume  contains  abstracts  of  a  series  of  lectures  given  to 
graduate  students  in  electrical  engineering  at  Union  College.  It 
is  primarily  intended  to  prepare  the  student  to  understand  and 
to  deal  mathematically  with  phenomena  which  are  incidental  to 
abnormal  or  transient  conditions  in  electric  circuits. 

The  first  part  is  practically  a  reprint  of  a  series  of  articles 
published  by  the  author  some  years  ago  in  the  General  Electric 
Review.  These  cover  the  simple  transients  in  circuits  containing 
concentrated  inductance,  capacity,  and  resistance,  which  have 
been  treated  by  many  authors,  notably  by  BEDELL  AND  CREHORE 
in  their  "  Alternating  Currents,"  published  1893. 

The  second  part  deals  with  the  somewhat  more  difficult  prob- 
lems of  transients  in  circuits  of  distributed  inductance,  capacity 
and  resistance.  These  were  treated  mathematically  very  fully 
almost  thirty  years  ago  by  Heaviside  in  a  series  of  papers  on 
" Electromagnetic  Theory/7  later  published  in  book  form.  In 
1909  Steinmetz's  " Transient  Phenomena"  appeared.  This 
book  covered  in  a  broad  sense  very  much  the  same  ground  as  that 
of  the  authors  given  above,  but  covered  it  in  an  essentially  differ- 
ent way;  introducing  for  the  first  time — asfa-r  as  the  author  knows 
— a  really  advanced  book  on  practical  electrical  engineering 
problems. 

The  third  part  of  the  book  deals  with  problems  in  electro- 
statics. These  again  have  been  very  fully  treated  almost  fifty 
years  ago  by  Maxwell  in  his  famous  books  on  "Electricity  and 
Magnetism. "  Since  that  time  a  large  number  of  papers  and  books 
have  appeared  on  the  subject,  notably  by  Heaviside,  Kelvin, 
Gray,  Jeans  and  Webster,  and  quite  recently  by  Coffin  in  his 
interesting  little  book  on  "Vector  Analysis." 

While  the  literature  on  this  phase  of  engineering  is  thus  very 
extensive,  it  has,  for  all  purposes,  been  closed  to  the  practical 
engineer  because  of  his  lack  of  sufficient  mathematical  knowledge. 
Dr.  W.  S.  Franklin  has,  however,  recently  published  a  number  of 
papers,  which  in  a  beautifully  simple  way  have  demonstrated 
that  these  advanced  problems  can  be  solved  with  simple 
mathematics. 


vi  PREFACE 

The  last  part  of  the  book  gives  an  outline  of  the  theory  oi 
electric  radiation.  The  mathematical  theory  was  again  given 
almost  fifty  years  ago  by  Maxwell.  Hertz's  verification  of 
Maxwell's  theoretical  work  given  twenty  years  later  and  pub- 
lished in  his  "Electric  Waves"  is  today  almost  the  last  word  in 
the  theory  of  wireless  transmission  of  energy.  Yet  it  would  be 
out  of  place  to  omit  a  reference  to  the  recent  excellent  papers  and 
books  by  Marconi,  Lodge,  Flemming,  Pierce,  Zenneck,  Cohen, 
Austin  and  a  score  of  others. 

It  is  evident  then  that  the  field  covered  in  this  volume  is  not 
new.  Nevertheless,  the  book  seems  justified  because  it  endeavors 
to  give  the  theory  in  a  way  comprehensible  to  students  who  have 
had  only  the  ordinary  undergraduate  course  in  electrical  engi- 
neering. It  is  hoped  that  the  volume  will  also  serve  a  useful 
purpose  in  bringing  to  the  attention  of  students  a  field  of  mathe- 
matics of  extreme  practical  importance  that  is  hardly  known 
to  them. 

The  author  is  greatly  indebted  to  one  of  his  graduate  stu- 
dents, MR.  M.  K.  TSEN,  who  not  only  examined  the  manuscript 
in  detail,  but  checked  and  elaborated  upon  the  theoretical  work. 
He  is  also  indebted  to  DR.  A.  S.  MCALLISTER,  who  kindly  criticized 
the  manuscript  prior  to  its  publication  and  offered  valuable 
suggestions. 


CONTENTS 

CHAPTER  PAGE 

INTRODUCTION 1 

PART  I 
TRANSIENT  PHENOMENA 

I.     CIRCUITS    CONTAINING    CONCENTRATED    INDUCTANCE    AND 

RESISTANCE 3 

II.     PROBLEMS  INVOLVING  MUTUAL  INDUCTANCE 33 

III.  CIRCUITS  OF  RESISTANCE  AND  VARIABLE  INDUCTANCE     ...  56 

IV.  CHARACTERISTICS  OF  CONDENSERS 68 

V.     A  CIRCUIT  CONTAINING  DISTRIBUTED  RESISTANCE  AND  IN- 
DUCTANCE       106 

VI.     CIRCUIT  CONTAINING  DISTRIBUTED  LEAKAGE — CONDUCTANCE 

AND  CAPACITY 110 

VII.     CIRCUIT      CONTAINING      DISTRIBUTED      RESISTANCE      AND 

CAPACITY 113 

VIII.     DISTRIBUTED  INDUCTANCE  AND  CAPACITY 127 

IX.     DISTRIBUTED     RESISTANCE — INDUCTANCE — LEAKAGE — CON- 
DUCTANCE AND  CAPACITY 143 

X.     PERMANENT  CONDITIONS  WHEN  ONE  OF  THE  FOUR  CONSTANTS, 

R,  L,  G,  AND  C  is  NEGLIGIBLE 148 

XL     DISTRIBUTION  OF  FLUX  OR  CURRENT  IN  A  CYLINDRICAL  OR 

FLAT  CONDUCTOR 150 

PART  II 
PROBLEMS  IN  ELECTRO-STATICS 

XII.     FUNDAMENTAL  LAWS 157 

XIII.     METHODS  OF  IMAGES,  APPLIED  TO  THE  PROBLEM  OF  POINT 

CHARGES  +  10  AND  —  5,  SEPARATED  5  CM 168 

XIV.  APPLICATION    OF    THE     POTENTIAL     FORMULA     V  =   2.* 

TO  SOME  MAGNETIC  PROBLEMS 180 

XV.  DIVERGENCE  OF  A  VECTOR,  POISSON  AND  LAPLACE  EQUATIONS.  186 
XVI.     LEGENDRE'S  FUNCTION 189 

XVII.     DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID 199 

XVIII.     CONCENTRIC  SPHERES 209 

XIX.     CYLINDRICAL  CONDUCTORS 218 

XX.     MUTUAL  AND  SELF-INDUCTION  OF  ELECTRO-STATIC  CHARGES 

OR  FLUXES — MAXWELL'S  COEFFICIENTS 232 

XXI.     TWO-CONDUCTOR  CABLE 237 

vii 


Vlll 


CONTENTS 


XXII.     THE  ELECTRO-STATIC  EFFECT  OF  A  THREE-PHASE  LINE  ON  AN 

ADJACENT  WIRE  OR  WIRES 249 

XXIII.  THE  CURL  OF  A  VECTOR 257 

XXIV.  THE  EQUATION  OF  THE  ELECTROMOTIVE  FORCE 260 

XXV.     SOLUTION  OF  ALTERNATING  CURRENT  IN  CYLINDRICAL  CON- 
DUCTOR— SKIN  EFFECT '. 271 

XXVI.     ELECTROMAGNETIC  RADIATION.  .   278 


APPENDIX    I:  PARTIAL  DIFFERENTIATION  .    . 
APPENDIX  II:  ELEMENTS  OF  VECTOR  ANALYSIS. 


319 
327 


INDEX 


331 


ELECTRCAL  ENIN-EERING 
ADVANCED  COURSE 

PART  I.  TRANSIENT  PHENOMENA 

CHAPTER  I 

CIRCUITS  CONTAINING  CONCENTRATED  INDUCT- 
ANCE AND  RESISTANCE 

The  study  of  transients  in  circuits  of  concentrated  inductance 
and  resistance  involves  as  a  rule  a  knowledge  of  the  solution  of 
linear  differential  equations  of  the  first  order. 

One  example  of  such  a  differential  equation  is: 

2+/i(*)»=/i(*)  (1) 

where  fi(x)  and  fz(x)  may  be  functions  of  x  or  constants,  but 
must  not  be  functions  of  y. 

For  the  sake  of  convenience  fi(x)  will  be  denoted  by  P  and 
fz(x)  by  Q.  P  and  Q  in  the  most  general  case  are  then  functions 
of  x  but  not  of  y.  Thus,  equation  (1)  becomes 

%  +  Pdv  =  Q  (2) 

A  solution  of  this  equation  can  be  obtained,  in  several  ways, 
all  of  which,  however,  involve  "educated  guesses." 
Let,  for  instance, 

y  =  uv  (3) 

where  u  and  v  are  unknown  functions  of  x,  which  will  be  deter- 
mined in  the  most  advantageous  way. 

Since  dy          dv          du 

y  =  uv,  -7-  =  u  -;  —  \~  v-j-  (4) 

1  dx          dx         dx 

Substituting  (3)  and  (4)  in  equation  (2), 
dv          du 


or 


Since  u  is  entirely  arbitrary,   this  expression   can   be  greatly 


4  ELECTRICAL  ENGINEERING 

simplified,  by  selecting  such  a  value  of  u  as  to  make  the  coefficient 
of  v  or  the  parenthesis  zero.     Therefore  let: 


, 

dx  u 

.'.  logu  =     -  fPdx  +  C. 

Since  the  simplest  possible  function  is  sought,  let  that  particu- 
lar one  be  chosen,  which  makes  C  =  zero.     Thus: 

log  u  -     -  fPdx, 
and  u  =  e-SPdx  (6) 

Substituting  now  this  value  in  (5),  there  is  obtained, 


.'.  v  =  fefpd*  Qdx  +  C. 

and  since  y  =  uv, 

y  =  t-fFd*  [f«fP**  Qdx  +  C]  (7) 

Special  cases: 
First.  —  Let  P  be  constant,  a;  and  Q  be  a  function  of  x 


and  y  =  e~ax  [feax  Qdx  +  C]  (8) 

Second.  —  Let  P  be  a  function  of  x,  but  Q  be  a  constant,  b. 


and  y  =  e-^Fdx  [bfefpd*  dx  +  C]  (9) 

Third.  —  Let  both  P  and  Q  be  constants,  a  and  6  respectively, 

dy 
' 


or>  _  _    i    n  -ax 

Fourth. — Let  P  be  a  function  of  x  and  Q  be  zero. 

and,  y  =  < 


INDUCTANCE  AND  RESISTANCE  5 

If  P  is  a  constant  a,  then  y  =  Ce~ax. 

Fifth. — Let  P  be  zero  and  Q  be  a  constant,  6, 

.   dy^b 

and,  y  =  bx  +  C.  (12) 

Two  useful  integrals  that  can,  of  course,  easily  be  solved  but 
will  frequently  appear  are  given  below  for  the  sake  of  convenience. 

eai  cos  ut  dt  =   -  -5  [o>  sin  ut  -\-  a  COS  co/1. 

a"  +  or 

rf  sin  at  dt  =  -          ~^\a  sin  co^  —  co  cos  co/1. 
a2  +  a;2 


/• 


A  study  will  now  be  made  of  the  equation  of  the  current  flowing 
in  such  circuit  when  the  impressed  e.m.f.  is  steady  and  also  when 
it  varies  with  time.  Referring  to  Fig.  1,  it  is  evident  that  the 
following  e.m.fs.  exist: 


FIG.  1. 

First,  the  impressed  e.m.f.,  E\ 

Second,  the  e.m.f.  consumed  by  the  resistance  =  ir; 

Third  the  e.m.f.  consumed  by  the  self-inductance  =  j^  --yr  or 


di 

Ldi> 


where  E  is  the  impressed  e.m.f.  in  volts, 

r  the  resistance  in  ohms, 

N  the  number  of  turns  of  the  coil, 

L  the  inductance  in  henrys  (assumed  constant), 

-77  the  rate  of  change  of  flux  at  a  particular  instant,  t,  and 
i  the  current  in  amperes  at  any  particular  instant. 


6  ELECTRICAL  ENGINEERING 

The  e.m.f.  consumed  by  self-inductance  can  be  expressed  as 
J*  or  L  -jT-  because  the  inductance  by  definition  is  : 

_AT0 
=  108i 

thus  Nd<t>  _      di^ 

10*dt  "      dt' 

The  equation  connecting  these  e.m.fs.  is  obviously: 


«'  *-*  +  L  (14) 

That  is,  at  any  instant  the  impressed  e.m.f.  E  is  numerically 
equal  to  the  e.m.f.  consumed  by  the  resistance  and  the  e.m.f. 
consumed  by  the  inductance.  Note  that  e.m.fs.  consumed  by 
but  not  e.m.fs.  of  resistance  and  self-induction  are  considered. 
The  latter  are: 

T  di 

—  ir  and  —  L  rr 
dt 

Equation  (14)  can  be  written: 

l+z'-f  f 

Compare  this  equation  with  (2)  and  note  that  P  =  T  and  Q  = 

LJ 

E 

-j-  are  constant  when  the  impressed  e.m.f.  is  constant  and  not 

function  of  t.     Thus  the  solution  is  found  in  equation  (10)  and 


*  /~v  1  '         1       "^  /  -f  t~*\ 

i  =  Ce     L    +  -  (16) 

The  integration  constant  C  is  determined  from  the  fact  that 
time  is  required  to  impart  energy,  that  is,  in  this  case  to  produce 
or  alter  a  magnetic  field. 

Before  the  switch  is  closed,  there  is  obviously  no  field  sur- 
rounding the  turns.  Shortly  after,  however,  there  is  a  current 
and  thus  a  field  which  appears  simultaneously  with  the  current. 

Thus  since  a  magnetic  field  can  not  be  produced  instantaneously, 
no  current  can  pass  at  the  very  first  instant.  Thus  for  t  =  0, 
i  =  0.  Therefore 

0  =  C.--L-  +?, 


INDUCTANCE  AND  RESISTANCE 

but  c°  =    1, 

therefore  0  =  C  +  ^,  and  C  =  -  ^; 

and  .       Ef 

I    =    —  11    —    € 

r  \ 

This  equation  shows,  that  as  t  increases,  the  current  increases, 
and  finally  reaches  a  value, 


Assume  now  that  after  the  current  has  reached  this  value,  the 
circuit  is  disconnected  from  the  generator,  and  at  the  same  instant 
short-circuited.  What  can  be  expected  to  happen? 

The  Dying  Away  of  a  Current  in  an  Inductive  Circuit. — Re- 
ferring to  Fig.  2,  since  the  coil  is  surrounded  by  a  magnetic  field, 
and  the  field  can  not  be  destroyed 
instantaneously,  and  since  the  mag- 
netic field  can  not  exist  without  a 
current,  it  is' evident  that  the  cur- 
rent can  not  disappear   instantan- 


eously, but  must  die  away  gradu-  FIG.  2. 

ally. 

Referring  to  equation  (15)  which  is  the  general  equation  of  the 
current  and  remembering  that  the  impressed  e.m.f.  E  is  zero, 
we  have: 


the  solution  of  which  has  been  shown  to  be: 

i  =  Ce'i*' 

To  determine  the  integration  constant,  it  is  remembered  that 
at  the  very  first  instant  when  t  =  O,  there  was  a  definite  current  / 
in  the  circuit. 

Thus,  i  =  I  when  t  =  0, 

which  substitued  above  gives: 

C  =  I, 
and  the  equation  of  the  decaying  current  becomes: 

=    -£  (is) 


8  ELECTRICAL  ENGINEERING 

If  dW  is  the  energy  delivered  during  a  short  interval  dt,  then 
the  rate  of  energy  supply,  or  power  is: 


_. 

dt 

The  practical  unit  of  power  is  the  watt,  which  is  work  done  at 
the  rate  of  1  joule  per  second.  At  any  instant  the  power  is  the 
product  of  the  instantaneous  values  of  e.m.f.  and  current. 

Thus  the  power  equation  corresponding  to  equation  (14)  is: 

Ei  =  i  X  ir  +  iXL~ 

=  i*r  +  Li  jt  (19) 

It  is  seen  from  this  equation  that  when  the  instantaneous 
value  of  the  current  is  i,  energy  is  being  dissipated  at  the  rate  of 
i2r  joules  per  sec.,  or  watts,  in  heat,  and  is  being  stored  in  the 

magnetic  field  at  the  rate  of  Li  y-  watts.     The  energy  that  has 

been  supplied  to  the  circuit  t  sec.  after  the  switch  is  closed  and 
the  current  started  is: 

Eidt     joules  (20) 


f 

Jo 


The  energy  dissipated  in  heat 

=  I  Prdt  (21) 

and  the  energy  stored  in  the  magnetic  field 

,  V 


(22) 

where  70  is  the  particular  value  of  i  when  the  time  is  t. 

In  almost  all  calculations  of  transient  phenomena,  the  ex- 
pression e~ax  is  met  with,  e  is  the  base  of  the  natural  logarithm. 
It  has  the  numerical  value  of  approximately  2.718.  To  calculate 
the  numerical  value  of  any  particular  expression,  the  ordinary 
logarithms  are  used.  Thus,  for  instance,  to  find  the  value  of 
y  =  c~°-2,  the  method  is  as  follows: 

log  y  =     -  0.2  log  e  =    -  0.2  X  0.434  =    -  0.0868 
+  0.9132  -  1, 

therefore  y  =  0.819, 

therefore 


INDUCTANCE  AND  RESISTANCE 


In  Fig.  3  are  shown  the  values  of  this  function  for  a  large  number 
of  values  of  the  exponents.  Since  this  curve  is  plotted  on 
rectangular  coordinate  paper,  it  is  rather  unsatisfactory  for 
small  values  of  the  exponent,  and  the  table  below  has  therefore 
been  worked  out. 


FIG.  3. 


X 

erx 

X 

e~x 

X 

e~x 

X 

e~x 

0.00 

1.0 

0.25 

0.78 

0.80 

0.449 

1.8 

0.165 

0.02 

0.98 

0.30 

0.741 

0.85 

0.427 

2.0 

0.135 

0.04 

0.96 

0.35 

0.705 

0.90 

0.407 

2.5 

0.084 

0.06 

0.942 

0.40 

0.67 

0.95 

0.387 

3.0 

0.05 

0.08 

0.923 

0.45 

0.638 

1.0 

0.368 

4.0 

0.018 

0.10 

0.905 

0.50 

0.607 

1.1 

0.333 

5.0 

0.0067 

0.12 

0.887 

0.55 

0.577 

1.2 

0.301 

6.0 

0.0025 

0.14 

0.870 

0.60 

0.549 

1.3 

0.273 

7.0 

0.0009 

0.16 

0.852 

0.65 

0.522 

1.4 

0.247 

8.0 

0.00034 

0.18 

0.835 

0.70 

0.497 

1.5 

0.202 

9.0 

0.00012 

0.20 

0.819 

0.75 

0.472 

1.8 

0.165 

10:0 

0.00004 

Example  No.  1. — A  coil  having  1000  turns  and  5  ohms  resist- 
ance is  connected  to  a  source  of  constant  potential  of  100  volts. 


10 


ELECTRICAL  ENGINEERING 


(a)  Show  at  what  rate  energy  is  being  delivered  to  the  entire 
circuit  and  to  the  resistance.     Show  at  what  rate  it  is  being 
stored  in  the  magnetic  field  as  the  current  is  increasing  after  the 
circuit  is  closed. 

(b)  What  is  the  rate  of  change  of  the  flux  when  the  current  is 
10  amp.? 

Referring  to  equation  (13), 

*    T?  (23) 


therefore  the  rate  of  energy  supply  to  the  entire  system  is  Ei 

watts, 

and 


Ei 


+ 

108    dt 


(24) 


The  current  will  begin  at  zero  value  and  finally  reach  a  value  of 

E 


i  =  I  =  —  =  20  amp. 


2000 

1800 

1600 

"1400 

£1200 

c 

fc  1000 


400 
200 


Rate  of  Energy  Supply  to  Inductive  Circuit 

Constants  of  Circuits 

e  =  100  Volts 

r  =      5  Ohms 

N  =  1000  Turns 


8  10  12 

Current  in  Amperes 

FIG.  4. 


14 


1G 


20 


The  rate  at  which  energy  is  dissipated  in  heat  is  i*r  and  the 
rate  at  which  energy  is  stored  in  the  magnetic  field  is: 

§  3?  -  w  -  *v  (25) 

The  three  curves  in  Fig.  4  show  these  rates. 

It  is  interesting  to  note  that  energy  is  being  stored  at  the 
greatest  rate  when  the  current  is  one-half  of  the  the  final  value. 
This  can  readily  be  proven  by  differentiation  of  equation  (25) 
and  equating  the  result  to  zero,  thus, 

E  -  2ir  =  0, 


INDUCTANCE  AND  RESISTANCE  11 


therefore  .       E_       / 

~  2r  ~  2* 

The  rate  of  change  of  the  flux  as  the  current  changes  is  obviously 
d<j>  _     E  -  ir 
~dt  ~  F  X  IF8' 

Therefore  when  the  current  is  10  amp.  the  rate  of  change  is 
5,000,000  lines  per  sec.  The  rate  of  change  is  greatest  at 
first  and  becomes  zero  when  the  current  reaches  its  final  value. 

The  determination  by  calculation  of  the  inductance  L  of  a 
circuit  is  usually  very  difficult,  in  fact  almost  impossible  except 
in  the  very  simplest  cases,  such  as  parallel  long  circular  con- 
ductors. Approximations  of  one  nature  or  another  have  almost 
always  to  be  resorted  to.  Usually  the  inductive  circuit  contains 
iron,  and  in  that  case  the  reluctance  (and  hence  the  inductance) 
is  not  constant  but  changes  with  the  degree  of  magnetization. 
Later  in  this  volume  the  effect  of  the  changing  inductance  in 
iron  circuits  will  be  considered,  but  at  present  it  shall  be  assumed 
that  L  is  a  constant  regardless  of  the  value  of  the  current. 

The  inductance  of  the  field  circuit  of  a  dynamo  can  readily  be 
determined  for  any  particular  field  current  by  experiment.  All 
that  is  needed  is  to  run  the  machine  at  some  speed  and  to  read 
the  voltage  and  field  current.  These  data  in  addition  to  those 
of  the  field  and  armature  windings  suffice.  By  definition, 

total  flux  X  turns 
current  X  108 

The  total  flux  per  pole  is  determined  from  the  voltage,  speed  and 
armature  winding.  Consider  a  10-kw.,  two-pole,  direct-current, 
110-volt  generator,  having  2.5  megalines  of  flux  per  pole,  and 
1500  field-turns  per  pole.  Assume  that  at  normal  voltage  its 
field  current  is  3  amp.  and  that  the  field  spools  are  connected  in 
series.  Thus 

T       2.5  X  106  X  1500  X  2       or 

L  =  3  x  1Q8  =  25  henrys. 

Example  No.  2. — Figs.  5  and  6  represent  the  direct-current 
generator  referred  to  above.  M  is  the  armature  and  F  the  field. 
If  a  voltmeter  of  11,000  ohms  resistance  is  connected  as  shown 
and  switch  S  is  opened  without  arc  when  the  field  current  in 
ammeter  A  is  3  amp.,  what  will  be  the  effect  on  the  voltmeter  and 
will  the  ammeter  and  voltmeter  read  in  the  same  direction  as 


12 


ELECTRICAL  ENGINEERING 


before  the  switch  was  opened?  Before  the  switch  is  opened  the 
current  flow  is  as  shown  in  Fig.  5.  As  the  switch  is  opened 
the  field  flux  can  not  die  away  instantaneously.  The  field  cur- 
rent therefore  can  not  die  away  instantaneously,  but  continues  to 
flow  through  the  only  available  path,  which  is  that  of  the  volt- 
meter. Since  the  resistance  of  the  voltmeter  is  11,000  ohms  it  is 
evident  that  the  voltage  across  the  instrument  becomes  at  the 
very  first  instant  very  high. 


FIG.  5. 


x~i  r 
i  t 


FIG.  6. 

It  tends  to  become  ir  =  3  X  11,000  =  33,000  volts. 

Thus  the  voltmeter  will  probably  burn  out  as  the  needle 
swings  to  the  opposite  side  of  the  scale.  The  ammeter  needle 
will  remain  stationary  for  the  first  instant  and  gradually  come 
down  to  zero. 

This  problem  gives  an  idea  of  the  nature  of  the  shock  that  is 
experienced  where  the  field  current  of  a  generator  is  carelessly 
interrupted  and  permitted  to  pass  through  a  person.  Depending 
upon  the  nature  of  the  contact  the  resistance  of  a  body  may  be 
from  1000  to  10,000  ohms.  If,  therefore,  a  person  touches  both 
sides  of  the  field  winding  when  the  field  circuit  is  interrupted, 
he  will  experience  a  very  severe  shock.  The  energy  stored  is 
usually  quite  considerable.  In  this  case  it  is  J^L/2  =  J£  X  25  X 
9  =  113  joules.  Since  1  joule  is  0.74  ft.-lb.,  the  energy  available 
is  84  ft.-lb.,  i.e.,  that  of  a  pound  weight  dropping  84  ft. 

It  may  bs  asked,  what  would  happen  if  the  voltmeter  were  not 
connected  across  the  field  winding?  Where  would  the  initial 
rush  of  current,  of  3  amp.  flow,  when  the  switch  was  opened? 

In  reality  it  is  impossible  to  open  the  field  switch  without  an 
arc;  therefore  the  current  can  not  be  interrupted  instantaneously. 


INDUCTANCE  AND  RESISTANCE 


13 


Furthermore  the  circuit  is  more  complex  than  assumed.  The 
field  winding  has  considerable  capacity  and  therefore  acts  as  if  it 
were  shunted  by  a  condenser.  A  portion  of  the  3  amp.  will 
therefore  flow  as  condenser  current,  but  a  large  portion  will 
appear  as  secondary  currents  in  the  iron  circuit  of  the  poles. 
This  phenomenon  will  be  understood  later  from  the  investigation 
of  circuits  having  mutual  inductance. 

The  problem  is  instructive  in  that  it  explains  frequent  burnout 
of  voltmeters,  and  in  that  it  teaches  that  the  voltmeter  should 
always  be  disconnected  before  the  switch  is  opened,  or  otherwise 
be  connected  on  the  armature  side  of  the  field  switch.  It  teaches 
also  that  in  opening  the  field  switch  a  relatively  low  resistance 
should  be  shunted  across  the  field  winding  to  prevent  high  vol- 
tage, and  finally  that  it  is  well  to  open  the  field  switch  slowly. 
The  importance  of  shunting  the  field  circuit  is  best  illustrated  by 
a  numerical  example. 

j  Example  No.  3  (Fig.  7). — Assume  that  the  field  circuit  having  a 
resistance  of  36.5  ohms  is  shunted  by  a  resistance  of  50  ohms,  and 
assume  again,  for  the  sake  of  simplicity,  that  the  field  current  of 


—it        \ 


r  =  36.5 
L=  25 


FIG.  7. 


3  amp.  is  Interrupted  without  arc  and  that  L  is  constant  at  25 
henrys.  The  total  resistance  in  the  circuit  is  then  50  +  36.5 
ohms  or  86.5  ohms.  Determine  the  current  in  the  field  winding 
and  the  shunted  resistance  and  the  voltage  across  the  field  coils 
which  is  the  same  as  the  voltage  across  the  resistance  after  the 
switch  is  opened. 

Referring  to  equation  (18) 


=  3e 


-3.46« 


For  t 

0 

0.05 

0.10 

0.20 

0.5 

1.0 

£-3.45< 

1 

0.84 

0.71 

0.50 

0.18 

0.03 

i 

3 

2.32 

2.13 

1.5 

0.54 

0.09 

iR 

150 

116.0 

107.0 

75.0 

27.0 

3.0 

14  ELECTRICAL  ENGINEERING 

It  is  seen  that  in  this  case  the  maximum  voltage  across  the  field 
coils,  which,  of  course,  occurs  at  the  moment  of  opening  the 
switch,  is  150  volts,  as  compared  with  33,000  when  the  voltmeter 
shunted  the  field  coils.  The  field  current  i  dies  away  very  rapidly. 
In  1  sec.  it  has  almost  disappeared.  The  energy  stored  in  the 
field  is  spent  in  heating  as  an  izr  loss. 

Example  No.  4. — Prove  that  in  discharging  an  inductive  circuit 
all  energy  stored  is  spent  in  heat. 

The  instantaneous  value  of  the  current  was  found  to  be: 

i  =  Ie~ll, 

therefore  the  energy  expended  in  heat  from  time  zero  to  infinite 
time  is: 

2r 


J"<--  /*« 

i*rdt  =  Pr\ 
=o  Jo 

[T  2r     -i  oo  T2r 

-*•'*].  --^ 


It  is  of  interest  to  study  the  rate  at  which  the  field  flux,  or  what 
is  equivalent,  the  field  current,  can  build  up  when  closing  the 
field  winding  on  a  constant-potential  busbar,  and  to  see  how  much 
more  rapidly  the  field  current  can  be  made  to  build  up  when  a 
considerable  resistance  is  inserted  in  series  with  the  field  coils. 

It  will  be  assumed  that  use  is  made  of  the  winding  described 
in  example  3,  that  is,  one  with  a  resistance  of  36.5  ohms  and 
inductance  of  25  henrys.  This  circuit  is  connected  to  a 
direct-current  busbar  having  a  constant  potential  of  110  volts. 
Referring  to  equation  (17), 

i  =        l  ~^     =3[1  -6-1-46']. 


The  lower  curve  in  Fig.  8  shows  the  result  of  this  calculation. 

If,  instead  of  exciting  the  winding  from  a  110-volt  main,  it  is 
connected  to  a  220-volt  circuit  and  sufficient  resistance  is  inserted 
in  series  to  keep  the  permanent  current  at  3  amp.,  the  rise  in 
current  will  be  more  rapid  than  in  the  first  case,  as  shown  in  the 
upper  curve  of  Fig.  8. 

There  is  an  interesting  mechanical  analogy  for  the  starting  or 
stopping  of  a  current  in  an  inductive  circuit. 

To  bring  a  train  up  to  speed  a  certain  force  is  necessary;  this 
force  must  overcome  the  friction  and  provide  the  necessary 
acceleration. 


INDUCTANCE  AND  RESISTANCE 


15 


Let  F  be  the  total  force  necessary,  and  fv  the  force  of  friction 
and  wind  resistance  which,  for  simplicity's  sake,  is  assumed  to 
be  proportional  to  the  velocity  v,  and  the  mass  M. 


Then 


F  =  fv  +  mass  X  acceleration 


or, 


dv        f  F^ 

dt  ~~  Mv  "  M' 


.2       .4 


.8      1.0      1.2     1.4     1.6      1.8     2.0     2.2     2.4    2.6 
Time  in  Seconds 

FlG.   8. 


If  the  drawbar  pull  F  as  well  as  the  coefficient  of  friction  /  be 
assumed  constant  during  acceleration, 

F  f  t 

-* 


where  C  is  the  integration  constant 


16  ELECTRICAL  ENGINEERING  . 

If  the  train  start  from  rest,  then  for  t  =  O,  v  =  O. 

F  F 

.'.  O  =  j  +  C,  or,  C         -  -, 

. -£[,-.- 

By  comparing  this  with  the  equation  for  the  starting  of  a  current 

Fr  r    ~~i 

in  an  inductive  circuit,  which  is,  i  =  — 1^1  —  e~L*  J,it  is  seen  that 

in  electrical  problems,  the  current  corresponds  to  Velocity,  the 
e.m.f.  to  the  mechanical  force,  the  ohmic  resistance  to  frictional 
resistance  and  the  inductance  to  the  mass. 

The  analogy  can  be  carried  further.  The  energy  stored  in  the 
magnetic  field,  %LI2,  corresponds  to  the  kinetic  energy  of  a 
moving  body,  %Mv2.  The  electromagnetic  momentum  LI  cor- 
responds to  the  mechanical  momentum  Mv,  etc. 

A  problem  involving  mechanical  as  well  as  electrical  transients 
will  next  be  considered. 

Find  the  equation  of  the  dying  away  of  the  field  current  in  a 
direct-current  self-excited  shunt  motor  disconnected  from  the 
circuit  and  permitted  to  decelerate  to  standstill. 

Let  the  moment  of  inertia  of  the  revolving  part  be  /.  Let  the 
full  speed  be  N  revolutions  per  second  corresponding  to  an  angu- 
lar velocity  of  0:0  radians  per  second.  Let  the  power  required  to 
run  the  motor  at  full  speed  but  at  no-load  be  P  hp.,  and  assume 
that  this  power  is  represented  by  friction  loss  in  the  brushes  and 
bearings,  which  is  a  very  close  approximation,  particularly  after 
a  few  seconds  of  deceleration,  when  the  core  loss  becomes  very 
small;  and  neglect  the  i2r  loss.  Assume  that  the  saturation  curve 
is  a  straight  line,  so  that  proportionality  exists  between  the  field 
current  and  the  flux. 

Let  the  normal  field  current  be  J0.  Let  the  normal  flux  per 
pole  corresponding  to  this  current  be  <£.  Let  the  armature  e.m.f. 
at  full  speed  and  flux  be  E,  and  the  total  field-circuit  resistance 
be  r,  and  let  the  motor  have  p  poles  and  each  field  spool  have  n 
turns. 

Mechanical  Calculations. — 1.  Determine  the  angular  velocity 
«o.  It  is,  a0  =  2irN. 

2.  Determine  the  friction  torque,  or  moment  Q.     We  have 

X  lb. 


INDUCTANCE  AND  RESISTANCE  17 


3.  Determine  the  stored  energy.     In  general  W  =  ^Mv2,  in 
the  case  of  a  revolving  wheel;  if  p  is  the  radius  of  gyration, 

W   =  %M  (2ir  pN)2  =  %Ia02, 
where  /  =  Mp2  =  moment  of  inertia. 

Thus,  with  revolving  masses,  /  takes  the  place  of  M,  and  a.  of  v. 
During  deceleration,  no  external  force  or  torque  is  applied. 
Thus, 

0  =  Q  +  I  -JT  =  torque  of  friction  +  torque  of  deceleration. 

(For  sake   of  simplicity  the  small   power  given   electrically  is 
neglected.) 

da  =      -jdt,or,a  =        j  t  +  C; 

for  t  =  0,  a.  =  a0.  .'.  C  =  a0- 

Q. 

.  .  a.  =  ao  —  -j  t. 

If  T  denotes  the  time  at  which  the  rotor  stops,  then  for  t  =  T, 
a  =  0. 

•     Ci  ^  T          '     T* 

=  a0~jT,     ..7   -  Q«O. 


And,  «o,_         /          t\ 

\l  ""  T) 


a  =  Oi0  —  jfj  t  =  «o 


Check  whether  all  energy  is  spent  at  t  =  T,  neglecting  the  sup- 
ply of  energy  from  the  diminution  of  magnetic  field  and  the  con- 
sumption of  energy  in  heat.  The  stored  energy  is  3^/a02. 

The  energy  consumed  by  friction  during  deceleration  is: 

I    force  X  vel.  dt  =    \    Q2*  Ndt  =    \    Qadt  =  Q 


substituting,  ^  _  / 

~  Q  Q:O' 

=  Q  L>|  «o  -  7  YQ2  a°2     =  ~'  Q'E'D' 

Electrical  Calculations.  —  If  the  field  current  remained  constant 
during  the  deceleration  (which  it  obviously  does  not),  the  arma- 


18  ELECTRICAL  ENGINEERING 

ture  voltage  at  speed  a  would  be:ei  =  —  E,  so  that  due  to  the  loss 


of  speed  alone,  the  armature  voltage  is  reduced  from  E  to—  E. 

If  the  field  current  is  reduced  from  /  to  i,  the  flux  is  reduced 
from  <l>  to  <p,  and  therefore  the  e.m.f  .  at  constant  speed  is  reduced 

in  the  proportion  y-- 

i  0 

Thus,  at  field  excitation  i  and  speed  a,  the  armature  voltage  is 


Jo  a 
but  E 

Jo~     ' 


But  the  relation  between  the  current  and  the  e.m.f.  is: 

di 


or, 


-  -t  - 

.-.  i  =  C€  ^  Lr      =  Ce  LT  2; 
for  <  =  0,  i  =  70.     .*.  C  =  J0. 

/.  i  =  /oe"2TZ. 
The  motor  stops,  when  t  =  T,  and  when  the  current  is: 

_rT 
10    =    J0€      ^ 

Remembering  the  equation   of  the  decaying  current  in   an 

_rT 

ordinary  inductive  circuit,  i  =  I0€  ^~,  it  is  evident  that  in  the 
case  of  a  decelerating  self-exciting  machine  the  current  does  not 
die  as  fast. 

After  the  motor  has  stopped,  the  current  obviously  dies  down 
according  to  the  law: 

•  -    •    ~I  (t~T) 


NOTE.  —  For  a  more  detailed  discussion  of  this  see  STEINMETZ'S  "Tran- 
sient Phenomena." 


INDUCTANCE  AND  RESISTANCE  19 

Verify  the  curve  (a)  in  Fig.  9  in  the  case  of  a  four-pole,  7.5-hp. 
motor  having  the  following  constants: 

P   =  4  70  =  2.75  amp. 

N  =  20  E  =  110 

7    =  0.25  r    =  40  ohms,  total 

$    =  1.5  megohms  per  pole  n    =  1000  per  pole 
P  =  0.72  hp. 

In  large  machines,  the  windage  loss  is  frequently  greater  than 
the  loss  in  the  bearings. 

The  windage  loss  may  be  assumed  to  be  proportional  to  the 
square  of  the  velocity.  In  other  words  the  torqu.e  necessary  to 
overcome  the  windage  is  proportional  to  the  speed. 

Assuming  again  that  the  electric  power  is  small  and  that  it  can 
be  neglected  then  the  equation  connecting  speed  and  time  during 
deceleration  becomes: 

0=  Qi  +  Qia  +  JF^' 

or,        da       Q<2  Qi 

~ 


when  .    r  Qi 

a  =  oto,  1  =  0.      .  .  O   =  «o  ~T  7T* 


"a=     ~  ft 
when 

i  =  T,  the  motor  stops,  a  =  0, 

- 


>2  «0  +  Qi 
-  y  T,  or,  log  -1- 

I     ,          ^2  «0  +  Q 


r,    a  a    . 

e  =  Eo  -  •  j-  =  —  ir, 

OLQ    IQ  Q!o 

a   .  T  di 

.  .  —  ir  =  IT  +  L  --1 


20  ELECTRICAL  ENGINEERING 

s+f  ('-=)-  «• 


ao 
which,  transformed,  becomes: 


where  4=14- 

v  —  _ 


and  £ 


•••  I +£(!-'•>-«. 


but 


when  £  =  0,  z  =  7o; 

rA 

.".  C    = 


If  the  problem  given  above  is  modified,  so  as  to  include  a 
windage  loss  at  full  speed  of  0.15  hp.  as  well  as  the  bearing  loss 
of  0.72  hp.,  the  constants  are: 

r  =  40,  I  =  0.25, 


L  =  2l€,  Q2  -  -  0.00525, 

«o  =  125.8,  Qi  =  3.16. 

T  becomes  9.04  sec.,  and 

A  =  5.71, 

^  =  10.60, 
K  =  0.021, 

•     „•    _.     7     (-10.60*  +504.6  -504.6e-»-°210 
.   .   &    —    i  06  • 

In  curve  6,  Fig.  9,  is  given  the  relation  between  the  current 
and  time  in  this  case. 


INDUCTANCE  AND  RESISTANCE 


21 


The  problems  considered  up  to  this  point  have  all  involved 
very  simple  integrations.  Frequently,  however,  this  is  not  the 
case,  and  to  solve  the  differential  equations,  it  is  necessary  to 
make  algebraic  transformations. 

The  most  important  of  these  transformations  is  to  separate 
fractions  into  partial  fractions. 


3.0 


2.5 


1,2.0 


1.0 


0.5 


(b 


5         6         7 
Time,  Seconds 

FlG.    9. 


10        11 


Almost  any  algebra  deals  with  this;  nevertheless  it  may  be 
opportune  to  refer  to  it  briefly  here,  although  it  is  suggested 
that  the  student's  memory  be  refreshed  by  reading,  for  instance, 
WILLSON'S  "  Advanced  Algebra,"  from  which  the  following  is 
largely  abstracted. 

If  FJ-\  is  a  fraction,  that  is,  the  numerator  is  of  lower  degree 

than  the  denominator. 

It  is  known  that  F(x)  can  always  be  expressed  as  the  product 
of  linear  factors,  which  are  not  necessarily  real. 

If  the  factors  are  real,  then  F(x)  can  be  expressed  as  the  product 
of  real  linear  and  quadratic  factors.  Two  cases  will  be  here 
considered. 

First. — No  factors  are  repeated. 

Example. — F(x)  =  (aiX  +  &i)  (a&  +  &2)  (a$x2  -f-  &3#  -f-  C3). 


Then 


Al 


-f 


F(x)       aix  +  bi       a2x  -[-  62       a3x2  +  63a;  + 


where  AI,  A2,  A3,  and  B3  are  constants,  which  can  readily  be 
found,  since  if  the  expression: 


22  ELECTRICAL  ENGINEERING 

aoxn  -f  aixn~l  +  a2xn~2  +  •  -  •  i  =  b0xn  +  to"-1  +  b2xn~2  +  •  •  • 
holds  for  all  values  of  x,  then  the  coefficients  of  like  powers  of  x 
must  be  equal,  thus  a0  =  &o,  «i  =  &i>  etc- 
Show  that 

*2l  1  2  5  4z  +  5 


Second. — Some  factors  of  the  denominator  are  repeated. 

F(x)  =  (aix  +  bi)2(a2x  +  62)  («3^2  +  to  +  c3)2 
Then 


+ 

Prove  that 

2x*  +  b  2  7  15 


(x  -  l)2(x  -  3)2       (x  -  I)2   '   2(x  -  1)    '    (x  -  3)2       2(x  -  3) 

The  application  of  this  transformation  is  found  in  any  transient 
phenomenon  in  which  disproportionality  between  magneto- 
motiveforce  and  resulting  flux  exists. 

As  an  example  the  condition  governing  the  self-excitation 
of  a  direct-current  shunt-wound  generator  will  be  considered. 
(For  a  more  detailed  discussion  see  STEINMETZ'S  " Transient 
Phenomena.") 

It  will  be  assumed  that  the  relation  between  the  flux  <p  and  the 
field  current  i  can  be  sufficiently  closely  represented  by  FROE- 
LICH'S  equation: 

9  -       ki  (1) 

1  +fcii 

Let  ec  be  the  e.m.f.  generated  per  megaline  of  flux  at  normal 
speed,  and  eQ  be  the  normal  e.m.f.  at  normal  flux  <f>0. 
Then  ec 


The  e.m.f.  e  corresponding  to  any  other  flux  <p  is: 

e  =  ec<p 
The  e.m.f.  consumed  by  the  resistance  is  ir. 

The  e.m.f.  consumed  by  the  changing  flux  is  77^  -£• 

1UU  dt 

if  (p  is  expressed  in  megalines 

and  n  is  the  total  number  of  turns  enclosing  the  flux. 


INDUCTANCE  AND  RESISTANCE  23 

Thus  .          n    d<p 

h  100  dt 
from  (1)  .  _      _JP__ 

K    ~~~    K\(f> 

<pr 
.  .  ec<p  =  T 7 — 

Separating  the  variables 


W/c/     —  /  j  i  v  —    \Jj\D 

n  e<f>(K  —  KHP)  —  <pr 

<p(eck  —  r  —  ecknp) 

To  integrate  this  the  fraction  is  broken  up  into  partial  fractions 
thus: 


100  dt       [A  B 

T- 


and  A  is  found  to  be 
and  B  is  found  to  be 


n  [<f>        ek  —  r  —  k^ec(p 

k 


eck  —  r 
k\r 


eck  —  r 
Integrating  each  term  we  get  after  a  slight  transformation 

~  -      ~  '08  *  - 


If  at  the  time  of  closing  the  field  circuit  the  remanent  flux 
is  <pr  and  the  corresponding  voltage  =  er  then  for  t  =  0,  <p  =  pr, 
e  =  er. 

When  C  is  determined  and  the  final  expression  becomes: 

n  FT     i      e          i      eck  —  r  —  kie  ~\ 

t  =  ln/>    /    i x    kec  log r  log  — ; —         —, —   • 

100ec(eck  —  r)  L          &  er  &  eck  —  r  —  kierl 

The  voltage  ultimately  reached  is  e  =  e$  when  t  =  °° . 


Thus  ,      eck  —  r  —  n/ioo 

loe;  — r—      — i —  =  —  oo  thus 

P    If    -  -•-    7*    — —    A%  -|P 

ecfc  —  r 
ec/c  —  r  —  K\en  =  0    or    eo  =  — ; • 


The  greatest  value  of  r  which  gives  a  positive  value  of  e0 

r  =  eck. 
The  condition  of  self-excitation  is  thus  r  ^  eck. 


24  ELECTRICAL  ENGINEERING 

Up  to  this  point,  the  problems  have  involved  inductive  circuits, 
on  which  a  direct-current  e.m.f.  has  been  impressed.  In  case  of 
alternating  current  the  impressed  e.m.f.  varies  from  instant  to 
instant  and,  while  a  harmonic  e.m.f.  is  usually  assumed,  fre- 
quently the  variation  represented  by  a  wave  is  much  more  com- 
plex. As  long,  however,  as  the  e.m.f.  is  obtained  from  a  dynamo 
of  symmetrical  poles,  no  matter  how  shaped,  the  e.m.f.  wave 
can  be  expressed  by  a  series  of  sine  functions  of  odd  frequencies. 

In  the  study  of  transient  phenomena  in  connection  with  alter- 
nating current,  the  equations  are  derived  for  the  fundamental 
wave  only,  that  is,  the  instantaneous  values  of  the  e.m.f.  are 
represented  by  e  =  E  sin  6. 

If  it  is  desired  to  know  the  result  with  distorted  waves,  the 
simplest  method  is  to  treat  each  harmonic  independently  and  to 
add  the  instantaneous  values  so  obtained.  If  the  effective 
value  is  desired  the  square  root  of  the  sum  of  the  squares  of  the 
effective  value  of  each  wave  should  be  taken. 

As  stated  previously,  the  instantaneous  value  of  the  e.m.f. 
is  generally  expressed  in  two  ways,  either  e  =  E  sin  coi  or  e  = 
E  sin  6,  or  the  expression  may  be  of  more  general  form  :  e  — 
E  sin  (at  -f  a)  and  e  =  E  sin  (6  +  a).  In  these  expressions, 
e  is  the  particular  value  of  the  e.m.f.  at  time  t,  or  at  phase  angle 
6,  and  E  is  the  maximum  value  of  the  e.m.f.  In  the  first  case, 
the  angle  ut  is  expressed  in  radians,  not  in  degrees,  w  is  the  an- 
gular velocity  =  2  TT/,  where  /  is  the  frequency.  The  relation 
between  radians  and  degrees  is  360°  =  2ir  radians,  thus  1  radian 


is  -77—    =   57.3°.     To   reduce   equation    e  =  E  sin  (cot  +  «)    to 
ZTT 

degrees  it  should  therefore  be  written  e  =  E  sin  (57.3  ut  +  a), 
where  in  all  cases  a  is  expressed  in  degrees,  as  is  customary.  To 
reduce  the  expression  to  radians  it  should  be  written 


Note  in  connection  with  this  that  in  the  expression,  y  =  sin  x, 
x  is  expressed  in  radians,  not  in  degrees.  To  bring  it  to  degrees 
the  equation  becomes  y  =  sin  57.3  x. 

In  the  development  the  value  of  the  sine  function 

Sin  x  =  x  -  I  +  I  -  ^  +  •  •  • 
x  is  again  expressed  in  radians. 


INDUCTANCE  AND  RESISTANCE 


25 


It  is  important  to  have  this  clearly  in  mind.  It  is  well  worth 
while  to  plot  some  curves  of  distorted  waves  from  equations  in- 
volving phase  angle  as  well  as  radians. 

Example  No.  5. — Verify  the  e.m.f.  wave  in  Fig.  10,  e  =  EI 
sin  ut  +  Es  sin  (3  ut  +  a)  for  El  =--  10,  E3  =  5,  a  =  30°  and 
the  frequency  25  cycles. 


1.2 

1.0 

8 

G 

4 

2 
0 
2 
4 
G 
8 
1.0 
1.2 

/ 

t 

f  ~\ 

\ 

— 

\ 



\ 

— 

/ 

/ 

\ 

— 

V 

7 

\ 

— 



/ 

\ 

/ 

€ 

0     .4 

0     .6 

0     .* 

0     1 

JO    1 

20    1 

0    160    U80    200    220     2 
Angle\in  Degrees 

10    2 

0     2 

50    3 

30    3 

20   3- 

0  ^3 



\ 

/ 



\ 

/^ 

/ 





\ 

/ 

/ 

, 





V^ 

/ 

\ 

t 



— 

\ 

^2 

.01 


.02 
Time  in  Seconds 

FIG.   10. 


.04 


Prove  by  integration  that  with  a  distorted  wave  : 
e  =  EI  sin  (ut  -f  «i)  +  E3  sin  (3  cot  +  «3) 

the  effective  value  is  e«//  =  "\/6i2e//  +  ^32e//« 

Thus  in  this  instance,  since  the  effective  value  of  the  funda- 

~F  10 

mental  wave  is  —  ^  =  —7=  =  7.07,  and  that  of  the  triple  har- 

monic is  —  7=  =  —  T=  =  3.53,    the   effective   value   of   the  wave 


V2 


recorded  by  a  voltmeter  is  e  =  \/7.072  +  3.532  =  7.9. 

Referring  to  Fig.  11:  prove  that  ammeter  A  when  placed  in  a 
circuit  carrying  10  amp.  direct  current,  8  amp.  60-cycle  current, 
and  5  amp.  125-cycle  current  reads  13.7  amp. 


26 


ELECTRICAL  ENGINEERING 


Harmonic  e.m.f.  Impressed  on  a  Circuit  of  Resistance  and 
Inductance  in  Series. — Let  time  be  counted  from  zero  value  of 
the  impressed  e.m.f.  and  let  the  e.m.f.  be  rising. 

Thus  e  =  E  sin  ut  where  e  is  the  instantaneous  value  of  the 
harmonic  e.m.f.  at  time  t.  E  is  the  maximum  value,  co  =  2  TT/, 
is  the  angular  velocity,  /  the  frequency,  r  the  resistance  and  L 
the  inductance  of  the  circuit. 

10  Amperes,  D.C. 


8  Amperes,  60  Cycle 


5  Amperes,  25  Cycle 


FIG.  11. 

If  i  is  the  instantaneous  value  of  the  current  when  the  e.m.f. 
is  e  then: 


e  —  E  sin  cot  =  ir  +  L  -77 
at 


or 


di 


r  .       E 


(27) 
(28) 


By  comparing  this  equation  with  equation  (2),  it  is  seen  that 

T  V 

T  =  P  and  j-  sin  ut  =  Q. 

P  is  not  a  function  of  the  independent  variable  t,  but  Q  depends 
thereon,  thus  the  solution  is  given  in  equation  (8). 
It  is 


(29) 


i  =  e    Ll  \   I  e+  L    Y  sm  ut  dt  +  C 
The  solution  of  this  equation  depends  upon  solving 

C  +r-tE  ,          E  r  ,rt 

I  €^L    jr  sm  wt  dt  =  Y~  I  €    L    sm  <*t  dt. 

E. 

j-  is  a  constant  and  can  be  left  out  of  consideration  at  present. 


INDUCTANCE  AND  RESISTANCE  27 

It  is  also  convenient  to  substitute  a  single  letter  for  Y'     Let  then 


L 

The  immediate  problem  then  is  to  solve  ftat  sin  co£  dt. 

An  integral  involving  exponentials  or  sine  functions  is  usually 
easy  to  solve,  because  the  differential  of  the  functions  are  similar 
to  the  functions. 

If  y  =  €ax  then  ^  =  aeax. 
dx 

Similarly  if  y  =  sin  ux,  then  -5-  =  co  cos  coz, 
or  if  y  =  cos  ux,  then     -  =  —  co  sin  cox. 


Thus  i    at  ,    _  j.    at 

€        Ui    —    —   € 


f 


and  /      .  ,  1  , 

sin  co*  dt  —  —  —  cos  cot  at. 

J  w 

Fortunately  for  the  engineer  there  are  only  very  few  methods  of 
integration  that  need  to  be  known.  One  of  these  is  "  Integration 
by  Parts." 

That  is:  fudv  =  uv  -  fvdu  (30) 

In  integral  ft*  sin  to*  dt,  let  u  =  eat  and  dv  =  sin  co*  dt. 

.'.  du  =  aeat  and  v  = cos  co* 

CO 

/.  yV*  sin  co*  dt  =  -  —  cos  cot  +    \  -  eat  cos  co*  dt        (31) 

CO  J     CO 

This  equation  is  indeed  more  complicated  than  the  original.  It 
is  evident,  however,  that  by  again  integrating  the  last  term  in  31, 
an  integral  results  which  contains  an  exponential  term  eat  and  a 
sine  term  instead  of  the  cosine  term.  Thus  the  final  expression 
will  contain  integrals  of  the  same  trigonometrical  and  exponential 
functions,  which  therefore  can  be  solved  directly.  However,  it 
is  somewhat  more  convenient  to  use  another  method. 
Referring  again  to  (30)  let  in  this  case: 

u  =  sin  co*  and  dv  =  eat  dt 

.'.  du  =  co  cos  co*  dt  and  v  =  -  eat 

a. 

.'.  ftat  sin  co*  dt  =  —  sin  co*  -  (  -  eat  cos  co*  dt        (32) 
a  J  a 


28  ELECTRICAL  ENGINEERING 

By  multiplying  31  by  -  and  32  by  —  and  adding  the  two  equa- 

OL  Ct> 

tions,  it  is  readily  seen  that 

./V"  sin  „(  dt  =  ^-.  (^  -  ^-')  (33) 

co-  +  a-  \      co  a      I 

T 

Substituting  «  =  r  and  remembering  that  xt  the  reactance  cor- 
responding to  the  inductance  L  is  2irfL  =  coL  and  that  the 
impedance  z  =  \/r2  +  x2. 

Then 


I  €  L  *  sin  wt  dt  =  e+ L     -^,[r  sin  coi  —  x  cos  coi] 


(34) 


Let  the  angle  of  lag  of  current  be  ft  thus 

tan  ft  =  -  and  r  =  z  cos  ft  (35) 

x  =  z  sin  ft  (36) 

Substituting  the  values  in  34: 

-t  +  r  t  L 

sin  coi  dt  =  e     L     -  sin  (cot  —  ft)  (37) 

z 

Referring  to  equation  29 

E  -Lt 

i  —  —  sin  (cot  —  ft)  +  Ce    L  (38) 

The  integration  constant  C  is  determined  from  the  particular 
problem  under  consideration. 

Assume  that  it  is  desired  to  find  the  value  of  the  current  at  any 
instant  after  the  switch  is  closed  and  the  alternating  e.m.f.  is 
impressed  upon  the  circuit,  and  that  the  switch  is  closed  at  time 
t  =  ib  when  the  instantaneous  value  of  the  e.m.f.  is  e  =  E 
sin  coii. 

Since,  as  has  previously  been  discussed,  it  is  impossible  to 
establish  a  magnetic  field  instantaneously,  the  current  can  not 
flow  at  the  first  instant.  Thus  for  t  =  t\,  i  =  0.  Substituting 
these  values  in  equation  38,  then: 

7?  r 

0  =  -  sin  (coii  —  ft)  +  Ce~Ltl, 

E    Z£l 
.'.  C  =  —  e  L     sin  (coii  —  /3), 

Substituting  this  in  (38) 

sin  (coi  —  ft)  —  e    •*•  lfa)sin  (uti  —  ft)  (39) 


i  =--  ~\s 


INDUCTANCE  AND  RESISTANCE  29 

It  is  often  convenient,  to  eliminate  t  entirely  from  the  expres- 
sion and  to  use  the  phase  angle  8  only  and  to  express  6  in  degrees. 
That  is,  the  e.m.f.  is  expressed  as  e  =  E  sin  6.     In  that  case 

B    =    ^t    =    27T/7. 

The  exponential  term  6~£a~'°  becomes €~^(fl"0l)  =  €~*((?~'l) 

T_     (9-fr) 

if  6  and  6\  are  expressed  in  radians  or  &     57.3    if  0  and  6\  are 
expressed  in  degrees. 

Thus  when  6  and  d\  represent  degrees 

Er  -r  <«-_*>  -] 

i  =  -I  sin  (e  -  (3)  -  e    *    57.3    sin  (0i  -  /3)  (40) 

The  equation  is,  however,  always  written 

i  =  | [sin  (0  -  i?)  -  €~*  "-'!)  sin  (0!  -  0)]  (41) 

and  it  is  understood  that  the  exponential  term  should   be  ex- 
pressed in  radians. 

Equation  (41)  can,  of  course,  be  derived  directly  by  using  the 
phase  angle  0  instead  of  cat. 

Thus  T  di 

E  sin  ir  =  er  +  L  -7- 

cfo 

may  be  written 

E  sin  0  -  ir  +  x  ^,  (42) 

where  x  is  the  reactance. 
Thus,  x  =  2-irffj  =  coL 

and  cot  =  6. 

.".  d#  =  codt  or  dt  =  - 

CO 

Prove  that  equation  41  is  the  solution  of 

di 

E  sin  6  —  ir  -\-  X~TC, 

The  exponential  term  in  equation  (41),  while  of  importance 
during  the  first  second  or  so,  ceases  to  affect  the  result  very 
shortly  after  the  switch  is  closed. 

Thus  the  equation  for  the  current  after  the  system  is  stable  is 

i  =  Ez  sin  (8  -  0)  (43) 

The  current  lags  behind  the  e.m.f.,  E  sin  0,  by  an  angle  0,  whose 

x 
tangent  is  -• 


30 


ELECTRICAL  ENGINEERING 


The  effective  value  of  the  e.m.f.  is 

E 

=  V5 

and  of  the  current  j  E 

~~  V2Z 

It  is  of  interest  to  note  that  the  transient  term  is  a  maximum 
when  sin  (Bl  -  /3)  =  1,  that  is  0i  -  0  =  90,  or  0i  =  90  +  /3. 

This  value  of  6\  also  gives  the  maximum  value  of  the  perma- 
nent current. 


1.4 
1.2 
1.0 
.8 
.6 
.4 
.2 
0 
-.2 
-.4 
-.6 
-.8 
-1.0 
-1.2 
-1.4 

I 

1 

/ 

z 

I 
/' 

6 

L   Impressed  Voltage 
'   Current  when  Swi 

)           "            " 

'                     ft                       t  »                    1 

v 

tch  is  Closed  at 

1 

\ 

— 

k 
& 

270° 
79  Giving  tli 
Value  of 

- 

= 
= 
- 

.','       ,','       ,' 

e  Permaaeu 
Current 

•^ 

> 

7T- 

~r 

s 

y 

S 

J 

Mill 

/ 

y 

\ 

1    LU= 

x 

A 

- 

s* 

/ 

S 

x 

/ 

^s^ 

OS. 

s^ 

~^^ 

\ 

/ 

// 

/ 

^ 

s\% 

/ 

\ 

k 

/ 

/ 

y 

/ 

/ 

s 

i: 

\\ 

\\ 

/ 

w/ 

\ 

\s 

i 

/ 

•y 

, 

/ 

/ 

\ 

X 

\\ 

^ 

0 

/ 

//' 

7* 

/ 

\ 

5 

/ 

^ 

/ 

/ 

/ 

/ 

\ 

\ 

\ 

\\ 

^ 

/ 

420^ 

M 

7 

\ 

\ 

j 

W_ 

iKT 

1201  1501  1 

5Q 

2 

0 

^ 

0 

^ 

\\\ 

,300°  330 

1 

3U 

390  l 

//// 

/^53 

4i 

0 

5 

0 

5 

w 

5 

0    6« 

An 

glei 

a  Degrees 

\ 

Y 

^U  |y 

1  \ 

y//// 

\ 

\ 

V 

V^k/ 

//' 

^// 

^ 

\ 

s 

\ 

_^^: 

-^// 

^ 

> 

X. 

•>s 

^K 

t 

'// 

^ 

X 

\ 

'  — 

^S 

s\ 

•"v 

? 

FIG.  12. 

The  exponential  term  is  zero,  that  is,  there  is  no  transient 
effect  if  #1  —  j8  =  6  or  #1  —  f3  or,  in  other  words,  if  the  circuit  is 
closed  at  such  a  time  as  would  give  zero  value  of  the  permanent 
current. 

Fig.  12  shows  a  series  of  such  transient  currents.  Each  curve 
corresponds  to  the  closing  of  the  switch  at  a  particular  value  0i 
of  the  phase  of  the  e.m.f. 

Thus,  for  instance,  curve  D  shows  the  starting  current  when 
the  e.m.f.  wave  has  a  phase  angle  of  +60°,  that  is,  when  0i  = 
60°.  These  curves  are  calculated  with  the  following  constants 

E  =  1  r  =  0.196  x  =  0.98. 

Problem  No.  6. — Check  some  curve  in  Fig.  12. 

It  is  of  interest  to  study  the  rate  at  which  energy  is  being  sup- 


INDUCTANCE  AND  RESISTANCE 


31 


plied  at  any  instant.     This  is  equal  to  the  product  of  the  e.m.f. 
and  the  current: 

p  =  d  =  Es'm  B  X  |[sin  (6  -  0)  -  e~x(0~dl}  sin  fa  -  0)]     (44) 

By  simple  transformations  the  equation  becomes 

-  cos  (20  -  0) 
P    : 


-^(0-00  -I 

-  e   *  sin  (0i  —  0)  sin  0J     (45) 


A -Impressed  Voltage 

-  Power  Input 
C-  Permanent  Power 


FIG.  13. 

The  first  term  in  equation  45  must  represent  the  power  at  any 
instant  after  the  conditions  have  become  stable.  This  power  is 
expressed  by 

E2 

P1  =  22  [cos  0  -  cos  (20  -  ft]  (46) 

It   consists   of   two   terms,    on   e  a  constant  term  ^  cos  /3,  the 

other  a  term  which  changes  with  double  frequency;  the  net 
result  of  which  over  a  complete  period  is  zero,  since  the  positive 
values  are  as  large  as  the  negative.  Thus  while  the  instantan- 
eous values  of  the  power  vary  from  instant  to  instant  and  may 


32  ELECTRICAL  ENGINEERING 

alternate  from  positive  to  negative  values  there  is  a  definite 
average  power  delivered,  which  is 

E2 
P  =  2z  cos  P 

The  exponential  part  of  the  power, 

P2  =  -  e~rx(e~9l)  sin  (0!  -  0)  sin  0  (47) 

is  gradually  decreasing  in  magnitude  as  well  as  oscillating  at 
normal  frequency. 

In  Fig.  13  are  given  three  curves;  the  first,  A,  is  the  wave  of 
the  impressed  etm.f.;  the  second,  B,  the  power  input;  and  the 
third,  (7,  the  power  curve  after  conditions  are  stable.  These 
curves  have  been  based  upon  the  constants  given  in  problem  6 
and  are  well  worth  reproducing  by  calculation. 

The  curves  show  that  during  the  transient  period  the  instant 
of  maximum  power  is  practically  the  same  as  that  for  permanent 
condition.  They  also  show  that  the  first  rush  of  power  is  greater 
than  that  which  corresponds  to  permanent  condition,  the  reason 
being  that  the  change  of  flux  during  the  first  part  of  the  cycle 
is  greater  than  during  the  corresponding  time  under  stable 
condition. 


CHAPTER  II 
PROBLEMS  INVOLVING  MUTUAL  INDUCTANCE 

Up  to  this  point  the  problems  considered  have  dealt  with  cir- 
cuits of  inductance  and  resistance  only.  However,  in  many 
circuits  of  commercial  interest  there  are  secondary  circuits  which 
are  more  or  less  closely  coupled  with  the  primary,  and  which 
influence  the  former  materially.  As  instances  of  such  circuits 
may  be  given  the  secondary  winding  of  a  transformer,  the  eddy 
currents  in  pole  pieces  of  generators  and  motors,  induced  cur- 
rents in  telephone  lines  running  parallel  to  transmission  lines,  etc. 

Sometimes  the  secondary  circuits  carry  currents  by  virtue  of 
impressed  e.m.fs.,  but  frequently  the  currents  are  the  result  of 
the  action  of  the  primary  currents.  With  a  change  of  primary 
current  obviously  there  is  a  change  of  the  flux  produced  by  the 
current  and  if  this  flux  interlinks  with  the  second  circuit,  e.m.fs. 
are  induced  therein,  the  values  of  which  become  higher  as  the 
interlinkage  becomes  more  nearly  perfect.  While  it  is  impossible 
to  arrange  two  circuits  so  that  all  flux  interlinking  one  will  also 
interlink  the  other,  the  condition  can  be  approached  reasonably 
close  under  the  most  favorable  conditions. 

The  limiting  case  is,  of  course,  perfect  mutual  induction,  which 
condition  will  therefore  first  be  considered  briefly. 

Two  Coils  of  Perfect  Mutual  Inductance. — Assume  then  that 
it  is  possible  to  place  two  coils  so  close  together  that  there  is  no 
leakage  flux  between  them,  that  is,  so  that  all  flux  that  surrounds 
one  coil  also  surrounds  the  other.  Let  the  first  coil,  the  primary 
coil,  have  NI  turns  and  r\  ohms  resistance,  and  the  secondary 
coil  N%  turns  and  r2  ohms  resistance.  Determine  the  open-circuit 
voltage  of  the  second  winding.  When  the  first  is  connected  to  a 
source  of  constant  potential  E,  we  have  obviously: 

F         »'r 

r 

The  rate  of  change  of  flux  is  thus 

d4>       E  —  i 
dt   ~~     A 
33 


34  ELECTRICAL  ENGINEERING 

Therefore  the  voltage  of  the  second  coil  e2  is 

Nz  d4>  N2 

~WTt~     ~Nl(1 

At  the  instant  of  starting,  when  ii  is  zero,  the  secondary  voltage 
is  e2  =  -  -  -rp  E,  that  is,  it  is  proportional  to  the  ratio  of  turns. 

7^ 

When  the  primary  current  reaches  its  constant  value  J0  = 

the  secondary  voltage  e2  is  zero.  If  the  secondary  winding  has 
more  turns  than  the  primary,  then  at  first  the  secondary  voltage 
is  higher  than  the  impressed  voltage.  It  decreases  rapidly, 
however,  and  soon  becomes  zero. 

Prove  that  the  two  voltages  are  equal  numerically  when 


Assume  that  two  coils,  which,  when  considered  alone,  have 
resistances  and  inductances  of  r\,  r%  and  LI,  L2,  respectively,  are 
placed  so  close  together  that  there  is  perfect  mutual  inductance 
between  them  (which  of  course  is  in  reality  impossible).  Find 
the  open-circuit  voltage  of  the  second  coil  if  the  first  coil  is 
connected  to  a  source  of  constant  potential. 

In  the  primary  we  have: 

E  =  ifl  +  L,  J- 

The   counter  e.m.f.  of   self-induction   of   the   primary   coil  is 
-  LI  -j7»  and  thus  the  voltage  of  the  second  coil  is 
N2      dii 


Check  the  values  of  the  primary  current  and  secondary  voltage 

as  given  in  full  lines  of  Fig.  14. 

for 

E  =  10  n  -  0.10  L!  =    2.5  NI  =  10 

r2  =  0.50  L2  =  10  N2  =  20 

In  the  case  referred  to  above  the  primary  current  will  rise  from 
zero  to  a  final  value  of  100  amp.,  while  the  secondary  voltage 
decreases  from  —20  volts  to  zero. 

If  when  the  primary  current  has  reached  its  final  value  the 
coil  is  suddenly  short-circuited,  what  will  the  primary  current 
and  secondary  voltage  be? 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    35 

The  primary  current  will  decrease  according  to  equation: 


Check  numerically  the  two  dotted  curves  in  Fig.  13. 

During  the  discharge  of  the  primary  the  number  of  coulombs 
are 


|     hdt  =    ( 

Jo  Jo 


=  100—  = 
ri 


2500  coulombs. 


1UU 

90 

80 
70 
S  60 

r 

<  40 
30 
20 
10 
0 
-2 
-4 
-6 

s-s 

•3-10 
p> 
-12 

-14 
-16 

-18 
20 

\ 

I 

E 

- 

\ 

_^, 

-i 

^ 

-^ 

s 

^ 

~- 

N 

/ 

•* 

* 

\ 

/ 

Two  Coils  of  Perfect  Mutual 
Inductance 
A-Primary  Current 
B-  Secondary  Voltage 

\ 

^/ 

\ 

/ 

- 

/ 

/ 

SN 

X 

/ 

\ 

^ 

/ 

> 

/ 

•» 

a 

/ 

1 

• 

0 

ri 

2 
m 

) 

V 

o 

or 

a 

s 

_4 

0 

50 

\ 

e^ 

n 

S 

2C 

! 

\ 

J- 

^ 

^ 

^ 

^Z 

? 

k^ 

^ 

/ 

/ 

/ 

\ 

/ 

/ 

/ 

~-~* 

/ 

x 

j 

x^ 

/ 

N 

/ 

*s 

/ 

•^ 

^ 

---, 

^ 

/ 

FIG.  14. 

Obviously,  when  connecting  the  primary  to  the  source  of  supply, 
the  number  of  coulombs  required  up  to  the  time  when  the  current 
becomes  stationary  is  infinite,  since  it  takes  infinite  time  for  the 
current  to  reach  this  value. 

Two  coils  of  resistances  and  inductances  of  ri,  r  and  LI  L  are 
connected  in  series  and  placed  so  close  together  that  it  is  assumed 
that  they  have  perfect  mutual  inductance.  What  will  be  the 
resultant  resistance  and  inductance  (a),  if  the  coils  are  wound 

3 


36 


ELECTRICAL  ENGINEERING 


in  the  same  direction;  (6),  if  the  coils  are  wound  in  opposite 
directions? 

The  inductance  of  an  air  coil  is  subject  to  rigid  mathematical 
determination,  but  the  complete  solution  is  very  cumbersome. 
However,  one  of  the  best  approximations,  that  of  BROOKS  and 
TURNER,  published  as  an  Engineering  Experiment  Station 
Bulletin  by  the  University  of  Illinois,  is: 


L  = 


cm- 


IQfr  +  12c  +  2R 


109(6  +  c  +  R)  ^  106  +  lOc  +  1.472 

X  0.51og10(lOO  + 


For  coils  which  are  not  extremely  thin  or  extremely  long,  this 
equation  becomes  approximately: 


L  =  7 


cm" 


(2) 


(b  +  c  + 
Where  L  is  expressed  in  henrys 

cm  =  centimeter  length  of  wire 

b  and  c  are  the  height  and  thickness  respectively  of  the  coil 
and  R  the  outside  radius,  all  in  cm. 


00000 

OO( 

00 

f 

ooooo 
ooooo 

00< 

oo< 

00 
00 

ft 

ooooo 

ooc 

00 

ooooo 

oo< 

oo 

i 

—R—       --> 

1*  '  ' 

FIG.  15. 

The  maximum  inductance  is  obtained  when  b  =  C  and  R 
(see  Fig.  15).     Then 


2C 


L  = 


0.27Cm 
C  X  109 


henrys. 


It  is  seen  that  the  inductance  is  proportional  to  the  square  of 
the  total  length  of  wire,  which  is,  of  course,  proportional  to  the 
turns.  Thus  the  inductance  is  proportional  to  the  square  of  the 
number  of  turns,  or 

L  =  KN\ 


(a)  Coils  in  the  same  direction. 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    37 

Let  N  be  the  number  of  turns  in  the  first  coil,  or, 

N  = 


and  NI  the  number  of  the  turns  in  the  second  coil,  or, 

JLl 

r 

The  total  number  of  turns  in  the  two  coils  when  considered  as 
one  coil  (which  is  permissible  when  perfect  mutual  inductance  is 
assumed)  is 


The  combined  inductance  is  then 

Lo  =  KN02  =  (VL  +  VLi)2  =  L  +  Li  +  2-v/LLL 

The  resistance  is  obviously  r0  =  r  +  r\. 

(b)  By  similar  reasoning  it  is  found  that  if  the  turns  are  in 
opposite  directions 

Lo  =  L  +  LI  —  2\/LLi  and  r0  =  r  +  ^i- 

From  the  above  it  is  evident  that  the  equation  for  the  starting 
current,  for  instance,  is: 


Two  Coils  of  Perfect  Mutual  Inductance  Connected  Simulta- 
neously to  Sources  of  Constant  e.m.f  s.  E  and  EI.  —  Let  r,  r\  and  L, 

LI  be  the  resistance  and  inductances  respectively,  and  assume 
that  the  circuits  are  closed  at  the  same  instant.  Assume  first 
that  the  coils  are  connected  in  the  same  direction,  that  is,  in  such 
a  way  that  the  permanent  current  in  both  coils  will  produce  mag- 
netic fields  of  the  same  polarity.  It  is  evident  that  in  this  case  the 
impressed  e.m.f.  has  to  overcome  not  only  the  resistance  and  in- 
ductance drop  due  to  the  current  in  the  coil,  but  also  the  e.m.f. 
which  by  transformer  action  is  induced  in  one  coil  by  the  change 
of  current  in  the  other. 

Consider  one  coil  alone,  for  instance  the  second   coil:     The 

counter  e.m.f.  of  this  coil  is  —  £i~5T'     If  it  has   N\  turns,   the 

voltage  per  turn  is  —  jj-  —rr'     Since  it  has  been   assumed  that 

/V  i   u/t 


38  ELECTRICAL  ENGINEERING 

there  is  no  leakage  field  between  the  two  coils,  it  is  evident  that 
this  same  voltage  per  turn  is  induced  in  the  first  coil  by  the  cur- 
rent in  the  second.  Thus  the  "  transformer  "  e.m.f.  in  the  first 

coil  having  N  turns  is  —  -^-  LI  -rr,  and  similarly  the  transformer 
e.m.f.  produced  in  the  second  coil  by  the  current  in  the  first  is 

Ni  T  di 

~WLdt 

But  N          IT. 


therefore  the  e.m.f.  in  the  first  coil  caused  by  the  mutual  flux  is 

L  T  dii  _     ^/rrdil' 
LI^       ~VLLldt 

Thus  it  is  seen  how,  when  the  mutual  inductance  usually  denoted 
by  M  is  perfect,  M  =  \/LLi.  In  reality  M  is  always  smaller 
than  \/LLi.  The  general  equation  dealing  with  e.m.fs.  consumed 
by  resistance,  inductance  and  mutual  inductance,  are  then 


To  solve  for  instance  i  the  following  transformation  is  conven- 
ient, multiplying  (3)  by  LI  and  (4)  by  —  M  and  add  the  equations 
so  obtained. 

li  is:  LiE  -  ME,  =  Liir  +  LLl  ~ 

at 

-  Mi.r  -  M2  ^  (5) 

Since  with  perfect  mutual  inductance 

M2  =  LLi  (6) 

Liir  -  LiE  +  MEi 


dii  _  LJT    <ti 
'  '  ~di  ~  Mrl  di' 
Substituting  this  in  equation  (3)  : 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    39 


or 

5j    hi 


dt    r  Lrl  +  Lir         Lrl  +  LIT 
Thus>  - 


(9) 

To  determine  the  integration  constant  C,  it  would  be  a  mistake 
to  assume  that  the  current  i  is  zero  when  t  =  0.  All  that  is 
known  is  that  the  combined  coil  can  not  be  surrounded  instan- 
taneously by  a  flux  —  it  takes  some  time  to  produce  or  alter  a 
magnetic  field,  because  a  transfer  of  energy  is  involved.  It  is 
possible  that  currents  will  flow  the  very  first  instant,  currents 
which  produce  m.m.fs.  of  equal  magnitude  but  in  opposite  direc- 
tion. One  particular  case  of  this  would  be  where  the  currents 
were  zero,  but  this  is  not  a  general  solution. 

What  is  known,  then,  is  that  no  flux  will  exist  the  first  instant. 
Thus  the  m.m.fs.  must  be  equal  and  opposite,  and  since  the  cross- 
section  of  the  magnetic  flux  and  the  direction  of  the  turns  are 
assumed  the  same  in  both  coils,  it  follows  that  for 


or  N   .  .    IL 

*i  =  -Niiu  "Ste 

Substituting  this  value  in  equation  (7) 

LiE  + 


(ID 


Mr, 
or  L\EJ  "— 


Lri  + 
for  t  =  0. 


i      E 
'  r  ~ 


_        MEir 
.  .  \j  —  — 


r(Lri 
ME 

77 

r(L 
Similarly  ii  is  found  to  be 


.       E       MEjr  +  LErt   ,     ™     t  ,10, 

I    =   ---  77—      —  =  -  r-  6      Lri+Lir  (12) 

r        r(Lrl  +  Lir) 


(13) 

i          r1(Lr1+L1r) 

Problem  No.  7.  —  Prove  by  complete  calculation  that  if  the 
terminals  of  the  second  coil  are  reversed  the  following  are  the 
equations  of  the  currents 


40 


ELECTRICAL  ENGINEERING 
E      LErl  - 


I    = 

r        r(Lrl 


r-  6      Ln+Lir 


Lri+Lir 


(14) 


(15) 


In  the  case  that  the  two  coils  are  excited  from  the  same  direct- 
current  busbars  when  E  =  E\  the  equations  become : 
For  coils  wound  in  the  same  direction: 

E 


+ 


•i+Lir    € 


m 


Ln+Lir 


For  coils  wound  in  opposite  direction: 

Lri  —  Mr m t 


ll 


1  — 


Lri 


e      Ln+Li 


100 


CO 


Starting  Currents 

Two  Coils  of  Perfect  Mutual  Inductance 
A  -  Coil  No.l 
B  -  Coil  No.2 

D  -  Coil  No'2  )  Wound  ™  °PP°site  Directions 


Wound  in  same  Direction 


(16) 
(17) 

(18) 
(19) 


FIG.  16. 

In  Fig.  16  are  given  four  curves  showing  the  currents  in  two  such 
coils  of  perfect  mutual  inductance,  having  the  following  constants : 

r    =    0.10 
ri  =    0.50 
L  =    2.5 
Li  =  10.0 
E  =  El  =  10  volts. 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    41 


It  is  assumed  that  they  are  connected  in  parallel  to  the  same 
source  of  direct  current  at  a  constant  potential  of  10  volts.  The 
full-drawn  curves  correspond  to  the  condition  in  which  the  turns 
are  in  the  same  direction;  the  dotted  curves  to  that  in  which  the 
turns  are  in  opposite  directions.  It  is  well  to  verify  these  curves 
by  calculation.  It  is  of  interest  to  note  from  the  full-drawn 
curves  that,  while  the  two  coils  are 
connected  to  the  same  source  of  con- 
stant potential,  during  the  first  few 
seconds  the  currents  actually  flow  in 
opposite  direction.  The  second  coil 
having  twice  as  many  turns  as  the 
first,  and  therefore  a  smaller  final 
value  of  current,  has  a  current  of 
negative  value  at  the  first  instant  of 

one-half  the  magnitude  of  the  current  in  the  first  coil.  Eventu- 
ally the  currents  become  positive  and  are  proportional  inversely 
as  the  ohmic  resistances. 

It  is  of  interest  to  deduce  the  equations  of  the  currents  in  the 
two  coils  when  the  first  is  connected  to  a  source  of  constant  poten- 
tial, and  the  second  is  short-circuited  upon  itself,  as  shown  dia- 
grammatically  in  Fig.  17. 


FIG.  17. 


CO 


40 


20 


Starting  Current 

Two  Coils  of  Perfect  Mutual  Inductance 
A   -    Coil  No.l 
B  -    Coil  No.2 


FIG.  18. 


Prove  that  with  the  coils  wound  in  the  same  direction: 

11-   -Lri 

r 


MEr 


Ln  +Lir 

t 


TTr*    1 


Ln+Lir 


(20) 
(21) 


42 


ELECTRICAL  ENGINEERING 


In  Fig.  18,  which  gives  the  values  of  the  currents,  it  is  of  interest 
to  note  that  the  current  in  the  second  coil,  under  this  condition, 
remains  negative  and  approaches  a  value  of  zero.  The  initial 
values  of  the  currents  are  twice  as  great  as  before.  Thus  the 
impedance  is  greatly  reduced,  as  would  be  expected  by  the  pres- 
ence of  the  short-circuited  winding. 

As  a  further  illustration  consider: 

Two  similar  coils  having  perfect  mutual  induction  and  calcu- 
late the  currents  in  the  two  coils  when  a  sine  wave  of  e.m.f.  is 
impressed  upon  one  coil  while  the  other  is  short-circuited. 

Referring  to  Fig.  19 


t ,  r,  x 


The  equations  evidently  become: 
di  di\ 


di 


FIG.  19. 


E   .  di       dii      E 

-  sin  e  and  ^-  -  ^  =  —  cos  0 


(22) 


Ex 
~ 


and 


dii 


E 


dB        2x  Zl  =  ~  2~r  COS 


i  =  -  e 


~  cos  0  d0  -f-  Ce 


-  ~  -  cos  (6  -  ?)  +  Ce    L9 

/JQ     I 


(23) 


where 


Z0  = 


2  +  (2x)  2. 

The  condition  determining  the  integration  constant  is   that 
when  the  switch  is  closed  no  flux  exists,  thus  i  —  —  ii. 
Let  then  the  switch  be  closed  when  0  =  0^ 
Thus  from  (22) 


and  from  (23) 


E 

-  2ii  =  —  sin  0! 


-yr  ~  cos  (0!  -  ^)  + 

£JQ     I 


(24) 
(25) 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    43 
From  (24)  and  (25) 

°  =  ^X  ^  [f5  COS  (Sl  ~~  ^  ~jr  sin 
'    > 


"""  -T1] ™ 

The  transient  term  disappears  when 

x  sin  0i 

•y-     COS      (01      —      (p)       = — 

Expanding  and  substituting  it  is  readily  seen  that  this  occurs 
when  tan  0i  =  — -,  that  is  when  0i  =  <p. 

The  transient  is  a  maximum  when: 

d    [x  .  sin 

-FT  COS    (0i   —  to) 


7* 

that  is  when  tan  0i  =  —  ^-     or 

0i  -  <?  -  90°. 

Limiting  Cases.  —  (a)  r  small  compared  with  2x.     :.<p  =  90° 
and  cos  (0  —  <p)  =  sin  0  and  cos  <p  =  0. 
The  transient  effect  is  greatest  when 

#    .  E 

.*.  ii  =  --  sin  0  and  %  =  ^-  sin  0  (27) 

(6)  r  large  compared  with  2#  when 

E  E 

il  ==  ~ir  and  Z<  ==  ~  2r  (28) 

When  dealing  with  commercial  problems  involving  mutual 
inductance  it  is  never  possible  except  as  a  first  approximation  to 
assume  perfect  mutual  inductance  M2  is  always  smaller  than  LL. 
In  that  case  the  solution  demands  differential  equations  of  the 
second  or  even  higher  order. 

Fortunately,  however,  the  solutions  of  these  equations  are  as 
a  rule  simple. 

The  majority  come  under  the  class  of  linear  differential  equa- 
tions with  constant  coefficient  or  they  are  of  the  types  given 
below.1 

1For  a  complete  discussion  see  '*A  Course  in  Mathematics," 
WOODS  AND  BAILEY,  vol.  II;  ''Differential  Equations,"  JOHNSON;  "Differ- 
ential Equations,"  MURRAY,  or  indeed  almost  any  book  on  differential 
equations. 


44  ELECTRICAL  ENGINEERING 

dzy 
First.        -r^2  =  f(x)  =  an  expression  that  is  a  function  of  x. 

Second.     ~r~2  =  fix,  -j-j  =  an  expression  that  is  a  function  of  x 

and  the  first  derivative  of  y  with 
respect  to  x. 


Third.      ^f2  = 

Fourth.     ~T\  =  f(y). 

Fifth.        Linear  differential  equations  -r—2  +  a  -j-  +  by  =  X. 

d^v  C13& 

First.         -j\  =  f(x)  =  a  function  of  the  independent  variable 
only.     By  integration  we  get: 

<fy  =  |/(a:)da;  +  d 
dx      J 

and  y  =  J '  J*f(x)dx2  +  CiX  +  C2. 

This  is  equally  true  for  ~[  =  f(x) 


Let  dy  d*y      rfp 

da;       P'         da;2  "  dx 

.'  .X  -j-  =  f(x,  p)  which  is  of  the  first. order  of  p  and  x. 

If  we  can  find  p  from  this  equation  then  we  can  find  y  because 
dy 
dx  =  V' 

dzy 


"**•   S-/(*Z) 


Let  dy  .     d2y  _  dp  _  dp  d/y  _     dp 

dx  ~~  ^        '  dx2  ~  dx  ~  dy  dx  ~  ^dy 

. ' .  X  p  -7-  =  f(y,  p)  which  equation  is  of  the  first  order  of  p  and  y. 
If  we  can  find  p  from  this  equation  we  can  find  y}  since  -^  =  p. 
Fourth. 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    45 

dii 
multiply  both  sides  by  2  -7-  dx 


Integrating  +  C 


==     = 

^fWv)dy  +  C 


+  c 

i. — Linear  differential  equations  of  the  second  order  with 
constant  coefficients.  If  the  coefficients  are  not  constant,  the 
solution  is  quite  complicated.  It  is  therefore  omitted  here.  As 
a  matter  of  fact  in  almost  all  problems  the  coefficients  are  con- 
stants. 

where  a  and  b  are  constants  and  X  is  either  a  function  of  z  or  a 
constant  or  zero. 

For  convenience  the  symbol  D  which  represents  the  differential 

operator  -r-  is  introduced. 

D  means  one  operation  of  differentiation  in  respect  to  xt  D2  two 
operations  of  differentiation  in  respect  to  x. 
Thus  equation  (29)  can  be  written 

D2y  +  aDy  +  by  =  X  or 
(D2  +  aD  +  b)y  =  X  (30) 

from  this  follows  that 

=  X 

y  ~  D2  +  aD  +  6' 

Obviously  this  does  not  mean  an  ordinary  fraction  but  is  simply 
a  symbol  to  express  the  solution  of  equation  (29).  The  denom- 
inator on  the  right-hand  side  is  not  a  number  but  is  an  operator 
which  has  a  definite  meaning,  so  for  instance  (D2  +  aD  +  b)emx 


46  ELECTRICAL  ENGINEERING 

really  is  the  sum  of  three  terms  the  first  of  which  for  instance  is 
obtained  by  differentiating  tmx  twice  with  respect  to  x,  the  second 
once  with  respect  to  x,  etc., 
The  expression  is: 

raV*  +  amemx  +  bemx 

Equation  (30)  can,  as  will  be  shown,  be  written 

(D  -  ml)(D  -  m^y  =  X  (31) 


(D  —  m2)y  means  -j-    -  mzy. 


Thus 


d  /dy  \  (dy 

(D  -  mi)(D  -  m2)y  =  ~ 


dzy  dy  dy 

dx-*-m*dx-m*dx 


(32) 

(Incidentally  it  is  seen  that  the  same  result  would  be  obtained  by 
the  simple  multiplication  of  (31)  treating  D  as  an  ordinary  quan- 
tity). Comparing  equations  (29)  and  (32)  it  is  seen  that 

-  (mi  +  mz)  =  a 
and  mim2  =  b. 


From  these  equations  mi  and  m2  can  be  expressed  in  terms  of 
a  and  b  which  are  known. 

A  slight  consideration  will  show  that  mi  and  mz  are  also  the 
roots  of  the  so-called  auxiliary  equation  m2  +  am  +  6  =  0,  which 
corresponds  to  the  so-called  complementary  function 


and  the  auxiliary  equation  corresponding  thereto 

m2  +  am  +  b  =  O 

\         l~2 

.*.  m  =  —  — h  *  \  -•    -  h 

2  -•  \4 

and  »    .      m-      ,  (33) 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    47 

The  rule  then  is  to  solve  the  so-called  auxiliary  equation 
m2  +  am  +  6  =  0 

and  find  the  values  of  mi  and  ra2 
Then  write, 

(D  -  ml)(D  -  m2)y  =  X. 
To  solve  for  y 

Let  u  =  (D  -  m2)y  (35) 

(D  -  mi}u  =  X 
and  du 

dx 

U    =   €«»1xyc-«1x  Xdx  _|_   ^m^ 

From  (35) 

An 

Cm  ~x  XT* 

l€     l      —    -A  i 


y  =  ee- 
Instance:  d2y          dy 


m2  -  3m  +  2  =  0 


m  =  1.5  ±  V2.25  -  2  =  1.5  +  0.5 

m2  =  1 


(Z)  -  1)(Z>  -  2)y  =  cosx 
(D  —  \}u  =  cos  x 

du 

-j  --  u  =  cos  x 

dx 

u  =  txj*e~x  cos  x  dx  +  Ciex 
but 

exj^e~x  cos  x  dx  =  %  (sin  x  —  cos  #)  by  simple  integration 

(D  -  2)y  =  y^  (sin  x  -  cos  x)  +  CV  =  Xl 
dy  (sin  s  -  cos  x) 

dx  ~  2y  ~  ~^T  Cie     =  Xl 

y  =  S*fe-**Xidx  +  C2e2*. 
The  three  integrals  involved  are  readily  solved  and  the  result 


s 


cos  a;          snx 
-         -  3  - 


In  many  cases  a  simpler  integration  is  obtained  by  the  following 
method  which  involves  the  breaking  up  of  a  fraction  in  partial 
fractions. 


48  ELECTRICAL  ENGINEERING 

It  is  known  from  algebra  that  if  the  equation  given  below 
holds  for  all  values  of  x  then  the  coefficients  of  the  like  powers  of 
x  are  equal. 

aQxn  +  aixn~l  +  an  =  b0xn  -f-  bixn~l  +  bn 
Thus  a0  =  b0 


0,2  =    z 
Equation  -j—2  +  a  -5  —  h  fa/  =  ^  can  be  written 


(D  -  mOCD  -  m,)?,  =  X  or  „  = 


But  it  is  known  from  algebra  that  the  fraction 

1  A  B 


(D  -  mi)  (D  -  w2)       D  -  mi       D  -  m2 
where  A  and  5  are  to  be  determined. 

_  1_  =  A(D  -m2)  +B(D  -mi) 

(D  -  mi)  (D  -  m2)  ~         (D  -  mi)  (D  -  m2) 

This  equation  shall  hold  for  all  values  of  D. 
Rearranging  the  equation  we  get: 

1  D(A  +  B)  -  (Am2  + 


(D  -  mi)  (D  -  m2)  ~  (D  -  mi)  (D  -  m2) 

On  account  of  the  identity  the  coefficient  for  D  must  be  zero  and 
the  constant  terms  must  be  equal,  thus 


and  Ani2  +  Bm\  =  —  1 

B  =  -  -  -  and  A  = 


m2  —  mi  m2  —  mi 


m2  —  mi        D  —  mi       D  —  m2 

_JLJr_^__        i    1 

mi  —  m2LZ>  —  mi       D  —  m2J 

Let  y  =  u  -\-  v 

u  =  ^^1D^1  (36) 

and  1  X 

~ 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    49 

Equation  (36)  written  out  is 

du  _  X 

dx 


u  = 
Similarly 


v  =  -  -  -ft-™**  Xdx 


The  general  solution  is  thus 

y  =  Ciemi*  +  C2em2*  +  - 

— 


,m2x 

"  <-*»**  Xdx. 


mi  —  mz 
When  X  is  a  constant  =  K 


The  solution  evidently  becomes: 

^  =  C^mx  +  Cucw«*  +  y- 

When  X  is  zero  the  equation  is  called  the  complementary  func 
tion  and  its  solution  is  found  by  making  K  =  0  in  the  above. 
The  solution  of  the  equation, 


is  y  =  Ciemlx  +  C2ew2a;. 

Before  leaving  the  subject  it  is  necessary  to  discuss  the  values 
of  mi  and  m2  which  involve  a  square  root  in  (33)  and  (34)  which 
might  be  real,  imaginary  or  zero. 

a2 
(a)  The  square  root  is  real,  that  is  -j  >  6. 

We  have  shown  then  that  mi  and  mz  depend  upon  the  auxiliary 

d^v          du 
equation  -j—2  +  a  -r-  +  by  =  0  and   that   the   solution   of   this 

equation  is, 


OL' 

(b)  the  square  root  is  imaginary,  that  is  b  >  - 


50  ELECTRICAL  ENGINEERING 


"fVlOTl  /  &        .       •  -4,    /•  O     \  /  d  .  >*    /.  Ct     \ 

,,  =  rJ-     +  '\*-i>  -4-c*(~2  -'\6--4> 

where  a  =  A/6  -   -j' 

But  e'ox  =  cos  ax  +  j  sin  ax 

. ' .  CV0*  =  Ci  cos  ax  -|-  Cij  sin  ax 

.  * .  C2e~;oa:  =  C2  cos  ax  —  Czj  sin  ax 
y  =  €~az  [(d  +  C2)  +  cos  ax  (Ci  -  C2)  sin  axj]. 

In  order  that  y  shall  be  real  it  is  necessary  that  C\  +  C2  and 
j  (Ci  —  C2)  shall  be  real,  in  other  words,  Ci  —  C2  must  be  im- 
aginary. 

Let  A  =  Ci  +  C2  and  B  =  j(d  -  C2) 

y  =  t~  2  [A  cos  ax  +  B  sin  ax]  (38) 

=  Cie~f  sin  (ax  +  C2)  (39) 

where 


and 

tan  C2  =  j}m 

(c)  the  square  root  is  zero,  that  is  -j-  =  6,  or  m\  =  m2 

This  is  not  a  complete  solution  of  the  complementary  function 
because  we  have  only  one  integration  constant. 
The  equation  can  obviously  be  written. 

(D  -  mi)2?/  =  0 

=  (D  -  mi}(D  -  mi)y  =  O  (40) 

Let 

(D  -  m,}y  =  v  (41) 

Then  40  becomes 

(D  -  mi)v  =  0 
or 

dv 

-; m\v  =»  O 

ax 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    51 

From  (41) 

(D  -  mi)y  = 
or 

fx  -  miy  = 

y    =    emla:[/Vwl*C 

or 

«i*  [ClX  +   C2]  (42) 


Two  Coils  Having  Resistance,  Self -inductance  and  Imperfect 
Mutual  Inductance  Constant  Impressed  e.m.f. — Let  the  constant 
e.m.f.  impressed  on  the  first  coil  be  E  and  that  on  the  second 
coil  EI.  Let  their  resistances  and  inductances  be  respectively 
r,  TI  and  L,  LI  and  let  M  <  LLj. 

It  follows  that 

E  =  ri  +  L  -j — \-  M  -r^  (43) 

and  dii  ,    ,,di 

E1  =  ^1rl  +  LI  -j — h  M  -r-  (44) 

Differentiate  (44). 


From  (43)  is  found 


Substitute   (46)   and   (47)   in   (45)  and   arrange  the  equation 
with  reference  to  the  derivatives. 

/72/  J7' 

/.  ^  (LLi  -  M2)  +  -ft  (Lir  +  Lri)  +  irr,  =  Er,        (48) 
Or  dH       vLir  +  Lri~\       di 


dt2   '    LLLi  -  M2J       dt    '   LLi  -  M2 

=  zzr--ir2  (49) 

E7 

Similarly  ii  =  ~  +  BI***  +  Brf-  (51) 

4 


52  ELECTRICAL  ENGINEERING 

where  mi  and  m%  are  the  roots  of  the  equation. 

Lir  -f  Lri  rri 

m   +  LL^W*m  +  LL^-W  -  ° 

(L,r  +  Ln)  +  VW-LrO'  +  Wrr, 


It  is  evident,  from  the  factors  under  the  square  root  sign,  that 
in  this  case  the  two  roots  are  real. 
Thus  the  solution  is 

L\r  -f-  Lri  - 


mi= 


2(LLi  - 


,„. 


and  Lir  +  Lri  ,     , 

2 


The  integration  constants  AI,  A  2  and  J3i,  7?2  are  readily  de- 
termined, since  in  this  case  (where  the  mutual  inductance  is  not 
perfect)  currents  can  not  flow  without  producing  some  flux,  and 
thus,  since  the  establishment  of  flux  requires  time,  the  currents 
can  not  appear  instantaneously. 
Therefore  at  t  =  0,  i  =  ii  =  0. 

Referring  to  (50)  and  denoting  the  final  current 

(where  /  ==  ^)  (56) 

by  0  =  7  +  ^,4-^2,  or  A2  =  -  (Al  +  7) 

we  get  i  =  I  +  Aie™i'  -  (Ai  +  I)em*  (57) 

and  ^  =  II  +  B^"1*  -  (Bl  +  7i)e^  (58) 

These  equations  still  contain  the  two  unknown  quantities 
A  2  and  B2.  To  determine  them,  multiply  (43)  by  LI  and  (44) 
by  -  M. 

L,E  =  L<ir  +  Lla       +  L,M  (59) 


-  ML1        -  M2 

di 
-  ME!  =  Liir  -  Mi^  +  (LL±  -  M2)  ^ 

~  LlE  +       (LLl  ~ 


The  value  of  -r  is  found  by  differentiating  (57)  and  the  value  of 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    53 

i  directly  from  equation  (57).  Substituting  these  values  in  (61) 
and  remembering  that  for  t  =  0,  i\  =  0,  the  integration  constant 
AI  is  found  to  be: 

Ltf  -  ME,  +  w,/(LLi  -  M2) 

-  -2 


.'.  A2  =  --  (A!  +  /)  (63) 

Similarly      n        LEl  -  ME  +  ma/i(LLi  -  M2) 
(mi  -  m2)   (LLi  -  M2) 


B2  =  -  (Bi  +  /O  (65) 

The  equations  of  the  currents  are  found  by  substituting  these 
constants  in  equations  (57)  and  (58).  They  are  so  long  and 
cumbersome,  however,  that  it  seems  unnecessary  to  insert  them 
in  this  text. 

Assume  that  the  two  coils  are  identical  and  wound  in  the 
same  direction,  and  are  connected  across  the  same  constant 
potential  busbars.  What  are  the  equations  of  the  currents? 

mi  and  mz  are  found  from  equations  (54)  and  (55). 

m>  =  -  L^M  (66) 

mz  =  -  LT_M  (67) 

AI  =  #1  is  found  from  equation  (62)  by  substituting  these  values. 

E  _ 

thus  A  2  =  B2  =  0. 

Referring  to  equation  (57) : 

(68) 


This  shows  that  the  mutual  inductance  acts  as  self-inductance. 
It  is  also  evident  that  if  the  two  coils  are  wound  in  opposite 
directions  the  circuit  is  almost  non-inductive.  It  would  be  non- 
inductive  if  M  =  L;  that  is,  with  perfect  mutual  inductance. 
It  is  of  particular  interest  to  study  the  relations  of  the  currents 
in  two  such  identical  windings  inductively  related  when  one  is 
supplied  with  current  from  a  source  of  constant  potential  and 
the  other  is  short-circuited. 


54         ELECTRICAL  ENGINEERING 

It  is  well  to  deduce  the  equations  from  the  two  general  ex- 
pressions: 


and  .  T    dii   .    ,f  di 

0  =  t,r,  +  L,  ^  +  M  % 

However,  it  is  evident  that  having  once  determined  the 
general  equations  (57),  (58),  (62),  (63),  (64)  and  (65),  it  is  possible 
to  give  the  equations  for  the  case  in  consideration  by  putting 
#1  =  0; 


that  is, 


T     _     I  _ 
=  n  = 


i  =  I  +  Aie"*  -.  (Ai  +  7)em««  (69) 

and  n  =  Bi€mi*  -  #iew2<  =  B!  (€™i<  -  €»*«)  (70) 


Referring  to  equation   (62)   and  substituting  equations   (66) 
and  (67) 


and 


(71) 

Referring  to  equation  (64)  and  making  similar  substitutions 
we  get 


M'  J 


(72) 


It  is  evident  that  these  equations  do  not  lend  themselves  to 
the  limiting  condition  M  =  L,  on  account  of  the  assumption 
made  in  determining  the  integration  constants;  that  is,  that  leak- 
age flux  exists  between  the  two  coils.  To  get  these  values, 
equations  (20)  and  (21)  should  be  used. 

In  Fig.  20  are  given  some  very  interesting  curves  which  show 
how  the  current  in  the  short-circuited  winding  depends  upon  the 
leakage  flux  between  the  windings.  These  curves  represent  the 
conditions  of  two  identical  coils  having  a  resistance  of  0.10  ohm 


PROBLEMS    INVOLVING    MUTUAL    INDUCTANCE    55 


and  an  inductance  of  2.5  henrys,  placed  at  various  distances 
apart  so  that  the  mutual  inductance  is  M  =  L  in  curve  a, 
M  =  0.9L  in  curve  b,  M  =  0.7L  in  curve  C,  M  =  0.5L  in 
curve  d,  and  M  =  0.1L  in  curve  e. 


-4 


2 


-1 


0     10     20    30    40     50 


70     80    90    100  110   120  130   140   150  160   170  180 
Time  in  Seconds 

FlG.   20. 


One  of  the  coils  is  connected  to  a  source  of  constant  potential, 
e  =  1  volt,  while  the  other  is  short-circuited. 

Prove  that  the  time  for  the  maximum  value  of  the  secondary 
current  is: 

Z^-M2         L  +  M 
Iog  L  -  M' 


CHAPTER  III 


10000 


8000 


CIRCUITS  OF  RESISTANCE  AND  VARIABLE  INDUCTANCE 

In  the  discussions  given  so  far  it  has  been  assumed  that  the 
inductance  L  has  been  constant.  In  almost  all  cases  of  interest 
to  engineers  this  is,  however,  not  the  case  because  almost  all 
magnetic  circuits  contain  iron,  and  the  permeability  of  iron  is  not 

constant  but  depends  upon 
the  magnetization.  In  other 
words  the  flux  produced  by  a 
given  current  is  not  pro- 
portional to  the  current. 
Fig.  21  gives  the  saturation 
curve  of  an  entirely  closed 
iron  magnetic  circuit,  as 
shown  in  Fig.  22.  It  is  the 
familiar  hysteresis  loop,  which 
shows  how  the  magnetism 
lags  behind  the  m.m.f.  pro- 
ducing it. 

This  particular  sample  has 
a  remnant  magnetism  of  7600 
lines  per  cm.2,  so  that  this 
density  corresponds  to  an 
exciting  current  of  0  amp. 
The  maximum  density  is 
10,000,  which  corresponds  to 
an  exciting  current  of  4.5 
FIG.  21.  amp.  If,  after  the  maximum 

density  is  reached,  the  current  is  gradually  reduced  the  rela- 
tion between  existing  current  and  density  is  found  in  curve  a. 
The  flux  does  not  disappear  until  the  current  is  2.6  amp.  in  op- 
posite direction  to  the  original  4.5  amp. 

If,  instead  of  being  entirely  of  iron,  the  magnetic  circuit  con- 
sisted partly  of  air  circuit  and  partly  iron  (Fig.  23),  the  influence 
of  the  air  circuit  would  as  a  rule  be  so  much  greater  than  that  of 

56 


-8000 


-10000 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  57 


FIG.  22. 


FIG.  23. 


the  iron  that  the  shape  of  the  saturation  curve  would  become 

materially  modified.     Thus  the  saturation  curve  of  a  dynamo, 

having     a     magnetic    circuit 

largely  of  iron  but  also  of  at 

least  a  small  air  gap,  can  be 

represented  by  a  set  of  curves 

similar   to   those  in   Fig.  24. 

If  the  air  circuit  is  very  small 

the  two  curves  corresponding 

to  a  and  b  in  Fig.  21  can  be  observed.     If  the  gap  is  reasonably 

large  the  two  curves  merge  into  one  as  shown  in  the  dotted  line. 

FROLICH  evolved  an  equation  of  such  a  saturation  curve  for  a 
magnetic  circuit  consisting  partly  of  iron  and  partly  of  an  air 
gap;  which,  modified  by  KENNELLY,  can  be  written  thus 

ki 

:  T+fo# 

where  <f>  is  the  flux  corresponding  to  an  exciting  current  of  i  amp. 


4  6 

Current  in  Amperes 

FIG.  24. 


If  the  number  of  turns  of  the  exciting  winding  is  known  then 
the  inductance  for  any  particular  value  of  current  i  can  be 
determined.  It  is 


N(f> 


-.  where  N  =  number  of  turns. 


:.L  = 


AT/clO-8 


i  +  kj 

The  general  equation  thus  becomes: 

d  ,  ^  di 


r 


NklQr 
1  +  kii 


dt 


iNkkMr*  di 
(1  -  kii)2  dt 


(73) 


58  ELECTRICAL  ENGINEERING 

To  solve  this  equation  the  variables  are  separated 

di 


or  /  «*_  _  10^ 

I        7        *\  9  "\7*7 


r 

J  (e-i 


The  integral  is  solved  by  breaking  up  the  fraction 

into  three  fractions 

C 


6  - 


~     h  i    i    T    •  ~h 


The  constants  can  be  readily  found  and  the  integration  carried 
out  without  the  slightest  difficulty. 

A  becomes  -= 
a2 

B  becomes  — r 
a2 

and   C  becomes  — • 
a 

The  result  is: 

Nk  rr          (1 


rr  l  +  /dz 

loe;  e  — — — : h 

La  e  —  ir 


108a  La  "  e  -  ir         1  +  M J 

where  a  =  r  -\-  eki. 

In  this  case  a  simple  solution  which  is  quite  accurate  is  obtained 
if  the  last  term  in  (73)  be  omitted  since  an  inspection  will  show 
it  to  be  small  as  compared  with  the  second  term. 
We  have  then 

di  _    .         JVMQ-8    di 

Separating  the  variables  we  get: 

di  W*t 

?  (74) 


(e  -  ir)  (1  +  fet)        Nk 
again  1  A  B 

A  +Be  +  i(Aki  -  Br) 


(e  -  ir)  (I  + 

Since  the  left-hand  member  does  not  contain  the  unknown  i 
and  since  the  constant  term  is  1,  we  get 
A  +  Be  =  1 
Aki  =  Br. 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  59 

Thus  A  =  _  _J1_  and  B  =       fei 

r  +  efci  r  +  eki 

The  intergral  is  thus  broken  up  into  two  simple  integrals 

r_     <**         r      ^     _ ,  C- 

J  (e-  ir)  (1  +  kj)       J  (r  +  efci)  (e  -  ir)  ^  J  ( 


r  +  «*i)(H- 
^ 

i  fci 


fci 


r  +  eki     *  e  —  ir 


If  it  is  desired  to  find  the  value  of  i  at  any  time  after  the  circuit 
is  closed  then  i  =  0  for  t  =  0 


(75) 
e  —  ^r        e 

The  curve  connecting  z  and  t  can  conveniently  be  obtained  by 
assuming  different  values  of  t  and  solving  for  the  left-hand  mem- 
ber of  the  equation.  The  value  of  i  can  then,  of  course,  be  easily 
determined. 

Curve  a  in  Fig.  25  shows  the  relation  between  the  exciting  cur- 
rent and  the  time  for  the  field  current  of  a  direct-current  generator 
having  the  following  constants: 

e  =  100  volts  =  voltage  impressed  on  the  field. 
r  =  100  ohms  =  field  resistance. 
TV  =  4000  =  total  number  of  field-turns  in  series. 

01  =  1  megaline  with  1  amp.  excitation. 

02  =  0.6  megaline  with  0.5  amp.  excitation. 
From  FROLICH'S  equation  follows  then: 

1  =  and  0.6  =  •'•  k  =  L5 


i  .i 

ifci  =  0.5. 


It  is  instructive  to  verify  this  curve. 

Curve  b  gives  the  corresponding  values  if  the  saturation  curve 
had  been  a  straight  line,  i.e.,  if  the  flux  were  1  megaline  for  1  amp. 
excitation,  and  0.5  megaline  for  0.5  amp.  excitation. 


60  ELECTRICAL  ENGINEERING 

In  that  case  the  inductance  L  would  be  constant  and  would  be 

4000  X  1,000,000 
L=  10'  X  1  " 

and   the   equation   e  =  ir  +  L  -r  would  be  100  =  lOOf  +  40  -^ 

in  which  case 

i=l-  e-2-5'  (76) 


^ 

f  •* 

7 

s 

/ 

^ 

^   ^ 

/ 

.0 

/ 

'    / 

/a 

7 

/ 

/ 

/ 

•g 

. 

/ 

2J  -6 

7 

/ 

3 

7 

g 

/ 

/ 

£  , 

// 

3  .4 
X 

; 

/ 

W 

A 

// 

K 

.2 

j 

1 

/ 

/ 

n 

f 

1.2 

Time  in  Seconds 

FIG.  25. 


2.0 


It  is  interesting  to  see  that  the  field  flux  builds  up  considerably 
slower  than  would  have  been  the  case  if  L  had  been  constant. 
The  reason  for  this  is  that,  while  at  the  final  value  i  =  1,  the 
inductance  is  the  same  in  both  cases,  for  all  smaller  values  of 
current  the  inductance  is  greater  because  the  flux  is  greater 
for  the  same  current.  When  the  saturation  can  not  be  expressed 
by  a  simple  equation,  there  is  no  better  method  than  to  calculate 
step  by  step. 

Let  Fig.  26  represent  such  a  saturation  curve.  Determine 
the  rise  of  current  when  a  constant  impressed  e.m.f.  of  100  volts 
is  impressed  on  a  coil  4000  turns  having  a  resistance  of  100  ohms. 


Thus 


N 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  61 


Using  differences  instead  of  differentials: 

N    A0 

6:=ir  +  WAi 
or  108  (e  -  ir)  M 

=  0.25  X  107  (1  -  i 


(77) 


(78) 


1.4 

1.2 

|  1-0 

I   *8 
c 

5  .4 
.2 
0 


.4  .8  1.2  1.6 

Current  in  Amperes 

FIG.  26. 


1.U 

.9 

.8 

.7 

a 

Q> 

3.6 

___ 

^~~f 

-^ 

•z~-~ 

" 

^ 

^ 

"' 

' 

^ 

-<^ 

'"* 

^. 

/ 

/ 

'/ 

/, 

' 

/> 

' 

/, 

/+ 

/f 

/ 

5  . 

p  .4 

/ 

/ 

/ 

/ 

w 

.3 
.2 

/,/ 

// 

/} 

/ 

-— 

/ 

/ 

} 

'/ 

"0       .1       .2      .3       .4      .5      .6      .7       .8      .9      1.0     1.1     1.2    1. 
Time  in  Seconds 

FIG.  27. 

If  the  values  of  current  are  determined  every  one  tenth  of  a 
second  A£  =  0.10. 

.'.  A<j>  =  0.25  X  106(1  -  i). 

The  actual  flux  at  any  time  is  of  course  SA<£  the  corresponding 
relation  as  obtained  by  the  use  of  the  differential  equation. 


62 


ELECTRICAL  ENGINEERING 


The  method  of  calculating  is  illustrated  in  the  table  given  below 
and  the  results  plotted  in  full-drawn  lines  in  Fig.  27. 


First  approximation 


Second  approximation 


t 

A*> 

SAy> 

i> 

i  -  1 

A? 

SA<p 

t 

0.10 

0.25 

0.25 

0.182 

0.818 

0.205 

0.205 

0.146 

0.20|    0.205 

0.41 

0.316 

0.684 

0.171 

0.376 

0.286 

0.30;    0.171 

0.547 

0.447 

0.553 

0.138 

0.514 

0.413 

0.40 

0.138 

0.652 

0.555 

0.445 

0.111 

0.625 

0.525 

0.50 

The  starting  of  an  alternating  current  in  an  inductive  circuit 
containing  iron  is  of  special  interest  since  almost  all  electrical 
devices  used  with  alternating  current  have  iron.  Unfortunately 
the  equations  are  very  complex  and  are  not  subject  to  solution, 
even  with  long  and  elaborate  treatment  by  series.  Even  in  the 
simplest  case,  when  the  saturation  curve  can  be  represented  by 
FROLICH'S  equation,  an  accurate  solution  is  not  possible,  although 
to  be  sure  it  is  not  difficult  to  bring  the  relation  into  the  form  of  a 
linear  differential  equation.  The  problem  in  that  case  can  be 
solved  as  far  as  a  mathematician  is  interested;  but  the  engineer, 
and  indeed  the  mathematician,  can  not  use  the  solution  for  any 
practical  purpose. 

To  illustrate  this  assume  that  an  alternating-current  e.m.f.  is 
impressed  on  a  magnetic  circuit  having  N  effective  turns  per 

ki 
phase,    and    a   saturation    curve   represented   by   <j>  =   ..    ,    ,    .• 

Assume  that  the  resistance  of  the  winding  is  r  ohms,  and  that  the 
impressed  e.m.f.  is  a  sine  wave.  At  any  instant  the  following 
relation  exists: 


E  sin  ut  =  ir  +  L  -r.  -\-  i  -r:' 
dt          at 


But  L  =  j-fTjp  where  N  =  number  of  turns 


1  i  is  read  off  the  saturation  curve  or  since  in  this  particular  case  a  satura- 
tion curve  which  can  be  expressed  by  FROLICH'S  equation  has  been  assumed 

for  the  relation  i  = 


1.5 


1.5 


1  +  0.5* 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  63 

again  neglecting  the  last  term. 


,  di 

.  .  E  sin  ut  =  ir  +  -  —  :—  :  —  .  -7- 
1  +  kii  dt 


where  a  =  JVfc  10~8 

substituting  for  1  -f-  k\i  =  —  ; 

y 


y  =  e~  ~T d' 

r  r  +  ki  E  sin  o><  dt 
J  1 1     i 


(8o) 

dto*  1    d 


2/i  i?/ 

r          r         a     dy 
k\y       ki       kiy  dt 

dy 
kiy  E  sin  ut  =  r  —  ry  —  a  ~~ 

dy          (r  +  fci  E  sin  cup  _  r_ 
dt+y~  a  'a 

r  +  ki  E  sin  ut 


--J-    .    „ 

i  E  sin  w<  d<  "] 

(82) 


the  solution  for  i  is  found  by  a  simple  substitution. 

Unfortunately,  however,  the  solution  is  not  in  a  simple  form 
and  can  not  be  simplified;  and  thus,  while  mathematically  the 
problem  is  solved,  practically  it  is  unsolved.  In  cases  like  this  it  is 
necessary  to  proceed  by  a  step-by-step  method. 

Consider  then  the  case  of  an  alternating-current  impressed 
upon  a  magnetic  structure  having  a  saturation  curve  of  any 
shape.  Let  it,  for  instance,  be  expressed  by 

ki 

=  r+i^i 

The  following  relation  exists  at  any  instant : 

E  sin  orf  =  ir  +      g  (83) 


64  ELECTRICAL  ENGINEERING 

where  r  is  the  resistance 

N 
/.  E  sin  cot  eft  =  ir  dt  +  j^  d<£  (84) 

PI  V  108  7>  ID8 

•*'  d0  =  sin  ^  d*  ~  ""  d*  (85) 


If,  with  full-load  effective  current  Ie  the  resistance  drop  is  p  per 
cent,  of  the  rated  voltage,  then 

7er  =  j^Q  —  r=,  and  for  any  other  value  of  /  as  i,  ir  =      ^   —  ^-  (86) 

'  2 


or  since                         ,  cos  coi  , 

d =  —  sin  cot  dt 

E  X  108r      ,  cos  co£          pidt 

d*  =    ~N — L~    ~u TooTT~ 

It  is  usually  more  convenient  in  alternating-current  problems  to 
introduce  6,  the  phase  angle,  instead  of  co£. 

In  that  case  0  =  ut  and  dt  =  — 

Referring  to,  (127) 

_  E  X  108  pin  6  dB  _         pidB       1 
0  ~      ~~N~  co         "  Ie  100  X/2COJ 


OT 

In  most  problems  E,  N,  $max  and  the  frequency  are  known, 
so  that  numerical  values  can  directly  be  substituted  in  the  above 
equation.  Since,  however,  there  is  a  relation  between  them,  one 
or  more  of  the  quantities  may  be  unknown. 

The  most  general  aspect  of  the  problem  is  given  by  eliminating 
the  numerical  value  of  the  impressed  voltage,  turns  and  fre- 
quency, and  specifying  the  maximum  value  of  the  flux  :  <j>  maximum 
=  $. 

We  have  the  following  well-known  relation  between  $,  N,  E 
and  co. 

2irf  3>N 


108  108 

.'.  •  -  (88) 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  65 

When  in  an  inductive  circuit  the  resistance  is  very  small  compared 
with  the  reactance  so  that  the  impressed  and  counter  e.m.f.  are 
equal  numerically. 

Substituting  this  value  in  (87) 


1.0 

/ 

NS 

/ 

\ 

/ 

\ 

/ 

\ 

\ 

' 

/ 

\ 

/ 
f 

^ 

/ 

v 

\ 

1 

l\ 

\ 

/ 

/ 

k 

^ 

1 

I 

\ 

\ 

\ 

1 

/ 

\ 

7 

/ 

i 

\ 

\ 

\ 

1 

/ 

/ 

/ 

/ 

\ 

\ 

/ 

/ 

\ 

\ 

/ 

/ 

\ 

\ 

7 

/ 
i 

\ 

\ 

\ 

0 

\ 

j 

c 

) 

$ 

3 

1( 

<? 

] 

50 

ll 

0 

2 

0 

2 

0 

270 

I 

'i 

irr 

e 

\ 

\_ 

t  * 

1 

\ 

\ 

£ 

1 

V 

[ 

I 

\ 

; 

L 

\ 

1 

V 

\ 

/ 

—  ,- 

/ 

\L 

\ 

i 

\ 

I 

\ 

/ 

/ 

\ 

/ 

/ 

s 

-1.0 

/ 

\ 

^.    ^ 

FIG.  28. 

Substituting  differences 

A0,  A  cos  0  and  A0  instead  of  differentials,  the  equation  becomes: 


66 


ELECTRICAL  ENGINEERING 


If  the  ratio  between  flux,  current  and  phase  angle  0  is  deter- 
mined every  10°  then  A0  =  10°  =  0.175  radians. 


cos  6  +  0.00124 


pj 


\ 


Numerical  Example.  —  Determine  by  "step-by-step"  method 
the  current  in  an  iron-clad  inductive  circuit  when  an  alternating- 
current  e.m.f.  is  impressed  thereon.  Assume  that  the  saturation 


\ 


\ 


ssaodtu y  ut  ^uooanQ 

FIG.  29 


curve  of  the  magnetic  circuit  is  represented  by  Fig.  26  and 
equation : 

<t>  =  i    ,'HKV  megalines. 


Assume  that  before  the  switch  is  closed  the  remanent  magnetism 
is  zero  as  is  practically  the  case  when  the  magnetic  circuit  con- 
tains an  air  gap.  Assume  further  that  under  normal  conditions 


CIRCUITS  OF  RESISTANCE  AND  INDUCTANCE  67 


of  operation  the  maximum  flux  is  1.4  megalines,  that  normal 
effective  current  is  1.7  amp.,  and  that  the  resistance  drop  is 
3.91  per  cent.  Then 

A</>  =   --  1.4  [A  cos  0  +  0.00286i] 
-  1.4  A  cos  0  -  0.004?; 

The  total  flux  is  obviously  ZA<£.  If  the  switch  is  closed  when 
the  e.m.f.  passes  through  zero  and  is  rising,  the  normal  flux  at 
that  instant  would  be  a  maximum  in  the  negative  direction  as 
shown  in  Fig.  29.  As  it  has  been  assumed  that  the  flux  really  is 
zero  it  is  evident  that  there  is  a  transient  stage  in  the  mag- 
netization before  permanent  condition  is  reached.  It  is  evident 
also  that  if  the  switch  were  closed  when  the  e.m.f.  was  a  maxi- 
mum no  transient  condition  would  result,  because  the  condition 
then  demands  zero  flux,  and  the  flux  is  assumed  to  be  zero. 
In  the  numerical  example  it  is  assumed  that  the  switch  is  closed 
when  the  e.m.f.  wave  passes  through  zero. 

The  method  of  using  the  above  equation  is  best  shown  by  the 
table  given  below. 


No.  1 

No.  2 

No.  3 

No.  4 

No.  5 

No.  6 

No.  7 

No.  8 

No.  9 

9 

Cos  6 

A  cos  0 

-  1.4  A  cos  0 

SA<£ 

i 

0.004i 

SA<» 

i 

0 

1.0 

0 

0 

0 

0 

0 

0 

0 

10 

0.98 

-0.02 

0.028 

0.028 

0.01884 

0.000075 

0.027925 

0.01875 

20 

0.94 

-0.04 

0.056 

0.0839 

0.0576 

0.00023 

0.08367 

0.0573 

30 

0.87 

-0.07 

0.098 

0.18167 

0.1289 

0.000516 

0.1811 

0.1288 

40 

0.77 

-0.10 

0.14 

0.32115 

0.2398 

0.000959 

0.3202 

0.2395 

50 

0.64 

-0.13 

0.182 

0.5020 

0.402 

0.001608 

0  .  5004 

0.4010 

60 

0.50 

-0.14 

0.196 

0.6960 

0.604 

0.002416 

0  .  6935 

0  .  6020 

70 

0.34 

-0.16 

0.224 

0.9175 

0.881 

0.00352 

0.9139 

0.8740 

80 

0.17 

-0.17 

0.238 

1.1519 

1.247 

0.00497 

1.1469 

1.238 

90 

0.00 

-0.17 

0.238 

1.385 

1.712 

0.00684 

1.3781 

1.699 

Column  No.  1,  phase  angle;  No.  2,  the  cosine  of  the  phase  angle;  No.  3,  difference  in  the 
value  of  the  cosine  between  two  successive  steps,  for  instance  cos  20°  — cos  10°;  No.  4  is 
self-explanatory;  No.  5,  first  approximation  of  the  flux  (sum  of  No.  8  of  the  preceding  line 
and  No.  4  on  the  line  under  consideration") ;  No.  6,  current  as  obtained  from  the  saturation 
curve  or  the  equation  if  such  is  given;  No.  7,  ohmic  drop;  No.  8,  second  approximation  to 
the  flux  which  takes  into  consideration  the  ohmic  drop  (the  algebraic  sum  of  No.  5  and 
No.  7) ;  No.  9,  current  corresponding  to  the  last  approximation  of  the  flux  column,  No.  8. 


CHAPTER  IV 
CHARACTERISTICS  OF  CONDENSERS 

The  charge  q  of  a  condenser  is  proportional  to  the  voltage; 
or  q  =  Ce,  where  C  is  the  capacity  the  value  of  which  depends 
upon  the  mechanical  construction,  dimensions,  etc.,  of  the  con- 
denser, and  6  is  the  voltage. 

The  charge  q  is  expressed  in  coulombs  or  ampere-seconds. 
Thus  the  charge  dq  given  in  a  time  dt  when  the  current  is  i  amp. 
is: 

dq  =  idt. 

The  capacity  is  expressed  in  farads,  a  very  large  unit;  so  large 
indeed  that  in  actual  practice  it  is  never  used.  The  capacities 
of  condensers  are  almost  always  given  in  microfarads,  that  is, 
in  a  unit  which  is  one-millionth  of  a  farad.  Nevertheless, 
in  all  formulae  involving  capacity,  C  stands  for  farads,  not 
microfarads  (m-f.)  unless  stated  to  the  contrary. 

To  give  an  idea  of  the  capacity  of  condensers  used  in  engi- 
neering, it  may  be  of  interest  to  know  that  the  ordinary  paraffine 
paper  and  tinfoil  500-volt  blocks  of  the  size  of  the  average 
text-book  have  a  capacity  from  1  to  2  m-f.  In  a  high-potential 
transmission  line  the  capacity  of  one  wire  against  neutral  is 
about  0.016  m-f.  per  mile.  The  capacity  of  underground 
cables  is  relatively  high.  Depending  upon  the  voltage  and 
type  of  cable,  etc.,  it  must  obviously  vary  much.  It  is  usually 
less  than  2  m-f.  per  mile  and  more  than  }/{ Q  m-f.  The  capacity 
of  an  ordinary  Ley  den  jar  is  extremely  small — a  very  small 
fraction  of  a  microfarad. 

The  fundamental  equations  for  the  condenser  are  as  stated 
above 

q  =  Ce  (1) 

and  dq  =  idt  (2) 

From  these  follow:                       q  /Q\ 

=  C 

dq  =  Cde  (4) 

and                                          .  =  |  (5) 

68 


CHARACTERISTICS  OF  CONDENSERS  69 

Substituting  (4)  in  (2) 

dp 

Cde  =  idl  ori  =  C  ~  (6) 

(it 

or  e,  the  voltage  across  the  condenser  =  -~J*idt  (7) 

The  rate  of  energy  supply  or  power  is  ei 
or  from  (6), 

Cde  de 


or  from  (3)  and  (9),  .  _  ?   .  _  q_  dq 

~  C  ^    "  C  dt 

The  energy  stored  in  a  condenser,  which  is  the  same  as  that 
required  to  charge  a  condenser  to  a  voltage  E  or  to  a  final 
charge  Q,  is  therefore  the  rate  of  energy  multiplied  by  the  time. 
It  is: 


f 

Jo 

Jo 


e 

o2 

/~n        ,  /•• 

or 


cedt  =  C  -  C 


2C 


Equations  (10)  and  (11)  are  obviously  identical,  since  at  any 
instant 

q  =  Ce  thus  for  e  =  E  when  q  =  Q 

Q  =  CE,  which,  substituted  in  (11),  gives 

CE2 


2C  2 

As  in  the  case  of  inductance,  the  calculation  of  the  capacity 
of  any  but  the  simplest  circuits  is  difficult.  It  will  be  dealt 
with  in  later  chapters. 

Of  particular  interest  to  engineers, 
however,  are  a  few  simple  forms  of  con- 
densers,  the  approximate  capacity  of 
which  are  given  by  equations  which  are  ^IG  39. 

well  known. 

Thus  the  capacity  between  parallel  plates,  Fig.  30  is: 

c  =  '  in  microfarads 


70 


ELECTRICAL  ENGINEERING 


where  K,  the  specific  inductive  capacity  is  approximately  1  for 
air,  2  for  paraffin  paper,  3  for  rubber,  5  for  mica  and  6  for  glass. 

A,  the  effective  area  is  given  in  sq.  cm.  and  d,  the  thickness 
of  the  dielectric,  in  centimeters. 

The  capacity  between  concentric  conductors  (Fig.  31)  is: 
0.0386LK 


C 


log 


in  microfarads 


io 


where  the  length  I  is  given  in  miles  of  cable,  K 
is  the  specific  inductive  capacity,  D  the  inside 
diameter  of  the  outside   conductor,   and  d  the 
diameter  of  the  inside  conductor.     This  is  the 
capacity    between     the     conductors,     not    the 
capacity  to  neutral  or  ground.     The  capacity  of 
one  conductor  1  mile  long  to  neutral  is  twice  as  great. 
The  capacity  between  transmission  lines  is: 

0.0386? 


FIG.  31. 


C  = 


in  microfarads 


where  I  is  expressed  in  miles  and  the  capacity  is  that  of  one  line 
against  neutral. D  is  the  distance  between  wires,  center  to  center, 
and  r  the  radius  of  wire.  The  charging  current  is  thus 

.        2irfCe 
106 

where  e  is  one-half  of  the  line  voltage  in  the  single-phase  system 
and  58  per  cent,  thereof  in  the  three-phase  system. 

Circuits  Containing  Concentrated  Capacity  and  Resistance 

Consider  at  first  the  case  of  a  constant  e.m.f.  E  impressed 
upon  a  circuit  of  resistance  r  and 
capacity  C,  Fig.  32.  After  the  cir- 
cuit is  established  a  current  flows 
and  energy  is  delivered  to  the  re- 
sistance and  the  condenser.  In  the 
resistance  heat  is  developed  and 
in  the  condenser  an  electrostatic 
field  is  produced.  The  energy  given  by  the  source  of  supply  of 
power  is  fEidt.  The  energy  supplied  to  the  resistance  is 

fftrdt 


FIG.  32. 


CHARACTERISTICS  OF  CONDENSERS  71 

and  the  energy  supplied  to  the  condenser 


Thus 


fEidt  =  fi*rdt  +    Cq  ~?  .  (12) 


Eidt  =  i*rdt  +  q  ~l 


Ei  =  izr  +  -„  Jj  which  is  the  power  equation          (13) 

and  ,-,        .          q    dq 

E  =  *r  +  Ci  dt 
or  substituting  for  dq  =  idt 

E  =  ir  +  -~  which  is  the  voltage  equation  (14) 

C 

Obviously  the  voltage  equation  could  have  been  derived  directly, 
since  ir  is  the  e.m.f.  consumed  by  the  resistance  and  ~  the  voltage 

across  the  condenser. 

The  condenser  voltage  is  thus  ei  =  E  —  ir;  but 


or  de,        I        =  E 

dt '  Cr?  "  *r 

Referring  to  equation^!  =  Ae    cr    -\-  E  (15) 


where  A   is  the  integration  constant:     The  current  is  readily 

CA    _JL<  A    _!, 

c    cv'  -e    c,  (16) 


r  .  „    ei 

found,  since  %  =  C  ~r 


The  charge  g  is  =  Cei  -  CAe~cr'  +  ^C.  (17) 

Special  cases: 

(a)  Condenser  charge. 

At  time  t  =  0,     e\  =  0. 

.'.  referring  to  equation  (15),  0  —  A  +  E     .'.  A  =  —  E 


(18) 
Referring  to  equation  (16) 

t--«-&'  (19) 


72 


ELECTRICAL  ENGINEERING 


Referring  to  equation  (17) 


EC\l  - 


(6)  Condenser  discharge. 

In  this  case  the  impressed  voltage  E 

Referring  to  equation  (15) 

i 

and  e0  =  A 
i  . 


(20) 


0  for  t  =  0,  ei  =  e0. 


=  At    crl 


(21) 
1  that  is  in  opposite  direction  to  charging  current  (22) 


2000  Ohms. 

•""*— -WV\A- 


FIG.  33. 


q  =  Ce0e~cr  (23) 

2oom.f.   Referring  to  the  e.m.f.  of  the  con- 
denser rather  than  to  the  impressed* 
e.m.f.,  the  current  becomes  positive 
since  the  discharge  current 

,-.-§-  -C^ 


dt 


dt 


_ 

—  C      Cr 


(24) 

Cr  *    "      '  r  -  (25) 

In  order  fully  to  understand  the  action  of  condensers  it  is  not 
sufficient  to  follow  the  equations  given  above,  but  it  is  essential 
and  indeed  necessary  to  figure  a  number  of  numerical  examples. 


h 

•2| 
||.05 

•S.2.04 

JU 

If02 

jfiOl 

§3 

oo  0 


.1     .2     »3     .4      .5     .6     .7     .8     .9     1.0    1.1 

Time  in  Seconds 

FIG.  34. 


90     1.8 
80     1.6 


6o|l.2| 


40^  .8 

0 

300  .6 

20  .4 

10  .2 

0  0 


For  this  reason  Figs.  33  and  34  are  given.     The  curves  shown 
there  should  be  checked  numerically  by  every  student.     They 


CHARACTERISTICS  OF  CONDENSERS  73 

are  calculated  under  the  assumption  that  a  constant  impressed 
e.m.f.  of  100  volts  is  impressed  on  a  circuit  of  2000  ohms  resist- 
ance and  200  m-f.  capacity,  as  shown  in  Fig.  33. 

An  interesting  problem  in  connection  with  the  charging  and 
discharging  of  condensers,  is  to  consider  the  flow  of  current  be- 
tween two  Ley  den  jars  of  different 
capacity  and  voltage  (Fig.  35)  .  The 
energy  stored  in  condenser  A  at  volt- 
age E  is  J^CE2.  The  energy  stored 
in  condenser  A  at  voltage  e  is 


The  energy  stored  in  condenser  B  at  FIG.  35. 

voltage  Ei  is  %CiEJ.     The  energy 

stored  in  condenser  B  at  voltage  e\  is  J^Ciei2.     While  current 

flows  between  the  two  condensers,  a  readjustment  of  energy  takes 

place. 

The  energy  equation  is  obviously: 

0.5CE2  +  0.5Ci#i2  -  0.5Ce2  -  O.SCi^2  =  fizrdt. 
By  differentiating  this  equation,  the  following  results: 

-  Cede  -  Cieidei  =  i*rdt  (26) 

As  it  is  assumed  that  the  voltage  of  A  is  higher  than  that  of  B, 
the  latter  being  charged;  thus 


where  e\  is  the  voltage  of  B  at  any  time.  Equation  (26)  contains 
three  variables,  e,  e1}  and  i,  which,  however,  are  dependent  upon 
each  other. 

At  any  instant  the  following  relation  exists  between  the  e.m.fs. 

e  =  ir  -f-  ei 

Thus  de_        di^       dei  _         d2^       dei 

dt   ~~    Tdt  ""  dt  dt2     ~  ~dt 

Substituting  in  (26) 

—  ClCir-rr  4-ei)  (Ctf-tt  +  -37)  —  dei-j^  =  Ci2r  (-57) 
\        eft          /  V        di2        d^  /  rfi  \  dt  I 

or 

6    ,   dei 


ei\      n     eilr  rei  j 
757  '   Cl  ~di(Cir~dt  +  ei 

or  dei          \  /      dei      rr     d2ei          deA 

)  lCl  ~di  +  CCir  W  +  c~dt)  =  ° 


74  ELECTRICAL  ENGINEERING 

Since  Cir  -57  +  61  can  not  be  zero 
at 


+  C-0  (29) 

Integrating  (157) 


or  dei          C  +  Ci         K 

Kl 


ei  =  K1  +  K2e-  (3D 

where  CCi 


/    —  /-?    i    /^ 

C  +  Ci 

The  integration  constants  KI  and  X2  are  determined  from  the 
initial  condition  that  for  t  =  o,  ei  =  EI  and  e  =  E 

.'.  Ei  =  KI  +  K2  or  K2  =  Ei  -  Kl 


i+  (E,  -Kfc     c» 


(33) 


ldt~        C0r" 
but  e  =  ir  +  e\ 

/.«--  ^(^i  -  Xi)e  "  ^  +  K! 
Co 

for  t  =  o    e  =  E 

:.  E  =-  ^(El  -  K,)  +  K 

Co 

.'.  K,  =  E,  +  ^ 
and  „  Co 


The  problem  can  be  solved  in  a  simpler  way  if  it  is  realized 
that  the  total  charge  in  the  system,  is  not  changed  after  the 
switch  is  closed.  Thus 

Qo  =  EC  +  E,     d  =  q  +  qi  (36) 

Where  q  and  qi  are  the  charges  at  any  time  in  jars  A  and  B 
respectively. 


CHARACTERISTICS  OF  CONDENSERS  75 

In  that  case  e  =  ir  +  e\\  or  since  q  =  eC  and  qi  =  e\C\, 


Assuming  E>Ei,  then  jar  A  is  being  discharged  thus 

dq 


or  g  i 

dt  '      Cdr  q  ~ 


where  ^  CC\ 

Co  ~  C  +  C 
for  J 


Since  condenser  B  is  being  charged 


K  —  EC  —  si  ,   si    —  Co(E  — 


Since  condenser  A  is  being  discharged 

,__*     +£z*j.-£,-. 

at  r 

The  voltage  across  condenser  A,  which  is  being  discharged  is 


for  t  =  o    e  =  E 

n 

„   ,   £,  ( 


76  ELECTRICAL  ENGINEERING 


=    O     —         =    O     —  —    C  (E   —   E  }     ~  ~Cor 

1 


TJ C0(E  —  EI 


1   E  -  E^ 

/"*  {s  QT< 

Ci        r 

i 


for  t  =  o, 


and  C      ,  _.      ,,  x  C 


With  a  slight  modification  of  this  equation  it  is  seen  that  for  t 
=  oo  the  final  voltage  between  the  coatings  of  the  Ley  den  jars  is 

E    =       Q° 

Numerical  example:  condenser  A  has  a  capacity  of  1  m-f.  and 
is  charged  to  1000  volts;  condenser  B  has  a  capacity  of  2  m-f. 
and  is  charged  to  500  volts;  the  resistance  is  10,000  ohms.  Find 
the  current  after  the  switch  is  closed. 

The  original  charge  in  A  is  then  1000  X   T™   =  0.001  cou- 

500  X  2 
lomb;  the  charge  in  B  is  — I-^G —  =  0.001  coulomb  also. 

E  -  EI  =  500 
2  X  106  2 


3  X  1012       3  X  106 
and  1 


C  +  Ci  =  3  X  10-6  c  Vc    =  0.667  c_^Ci  =  0.333 

•••  *' =  i<w--150' =  °-05€-150' 


CHARACTERISTICS  OF  CONDENSERS 


77 


e  =  1000  -  0.667  X  500(1  -  e-150<) 
=  500[2  -  0.667(1  -  e-1500] 

i  =  500  +  0.333  X  500  (1  -  €-1500 
=  500[1  +  0.333(1  -  e-160')]. 


14 
12 

,10    1000 
i 

I 

I  8  «  80°  S 

i6  >  600  1.060 
•S        <j 

|    4^1400-5.040 

|    2W*200  g.020 

>  £ 
000 

.140 

.120 

10    1000     .100 

I  « 

!    8  5  800  |  .08 

>         S 
i    6.SCOO-J  .06 

ir*        -^ 
^  4-» 

.    4  3  400  g  .04 
?  2      200  O  .02 

>  0         0          0 
-2    -200     -.02 
-4    -400 


\ 


Tim 


Case  No.2  Voltages  H 


-500 


)00 


.005 


.010  .015 

Time  in  Seconds 


FlG.   36. 


025 


.025 


For  t  =  oo ,  eQ  =  e0i  —  0.667  volts  which  is  the  final  voltage  of 
the  two  jars. 

Fig.  36  gives  the  result  of  these  calculations. 

Harmonic  E.m.f.  Impressed  upon  a  Circuit  of  Resistance  and 
Capacity  in  Series. — Let  e  =  E  sin  cot  be  the  impressed  e.m.f.,  r 
the  resistance,  C  the  capacity  and  q  the  charge  at  any  instant. 


78 


ELECTRICAL  ENGINEERING 


Then,  referring  to  Fig.  37, 

E  sin  wt  =  ir  -{- 
Differentiating 


=  ir  -\-  -^  \  i 


-    \  idt 


di 


u  cos  u>t  =  r-r  +  -^ 
di        i        Eos  cos  cot 


(1) 


E  Sin 


FIG.  37. 


Thus, 


_  i  t[  C  +-1-tEu  J 

i  =  e    Cr   \    I  €    Cr  —  cos  cotdt  +  K 

The  integration  is  readily  made  and  the  result  is: 


E 
i  =  -y  sin  (cot  +  j8)  +  -KV 


where 
and 


Z  =       r' 


a^r 

tan  /8  =  —  • 
r 


The  voltage  across  the  condenser  is: 


=  ~         sn 


fldt  + 


^  fe    c 


At  the  moment  of  closing  the  switch  ec  =  0. 
Thus  for  t  =  t1}  ec  =  0. 


(2) 


(3) 


thus, 


or 


t  =  |  [sin  (8 


cos 


cos 


(4) 


(5) 


CHARACTERISTICS  OF  CONDENSERS  79 

As  an  interesting  application  of  these  equations  and  the  corre- 
sponding equation  for  inductive  circuits,  consider  the  nature  of 
the  current  supplied  to  a  tuned  circuit,  Fig.  37,  when  the  resist- 
ance is  small. 

x  =    -  xCjZ  =  \/r2-}-  x'2  =  -\/r2-\-xc2 

The  line  current  at  any  time  is  the  sum  of  the  currents  in  the 
two  circuits. 

Q  (0  -  8)  +  sin  (0  +  8}  -  e  ~x(e~ei} 


sn    01  -  0    - 


=  ^|  2  sin  0  cos  j3  —  e      r    '     '  cos  (0i  —  /?) 

XC.  ""I 

r  e     '  cos  (0i  -f  )8)  (7) 

The  line  current  is  a  combination  of  a  sine  wave  of  form  -  — ~ — 
sin  0  and   two   exponential   or  logarithmic   curves.     Since   r  is 

^_    fa a   \ 

small  compared  with  x  or  xej  one  of  the  logarithmic  curves  e    * 

M  *^cfa         a 

sin  (0i  — ft)  dies  down  at  a  slow  rate,  whereas  the  other  -V  7l 

cos  (0i  -f-  (3)  dies  down  with  extreme  rapidity. 

x 
In  the  limiting  case  when  -  is  large  and  ft  therefore  approaches 

90°  the  permanent  term  disappears  since  cos  0  =  0, 
and  „       Er   -J-(g-gi)  xc  _xf(0_0,)    .        ~| 

0   =  '7?  I   €      *  COS  0i   H €       r  sin  0i 

ZL  r  J 

for  # 

r  =  OLto  =  ~~  cos  QI 

x 

Thus  the  interesting  situation  occurs  that  if  an  alternating 
voltage  is  impressed  on  a  tuned  circuit  as  shown  in  Fig.  37  and  if 
the  resistance  is  zero  the  line  current  is  a  steady  unidirectional 
current  having  the  value: 

.  _  E  cosd  i 
x 

If  the  switch  is  closed  when  0i  =  0,  then  the  direct  current  is  a 

maximum  and  is  — .     At  any  other  time  it  has  a  smaller  value. 
x 


80 


ELECTRICAL  ENGINEERING 


In  Fig.  38  is  shown  a  series  of  curves  which  illustrate  this  in 
the  case  where  the  resistance  is  considerable  and  the  circuit  is 
closed  when  0i  =  0.  The  constants  for  the  circuit  are: 

E  =  1,  r  =  0.05,  x  =  xc  =  1 

z'o  is  the  total  line  current,  the  dotted  sine  wave  is  the  impressed 
e.m.f.,  IA  the  current  in  the  condenser  circuit,  and  IR  the  current 
in  the  inductive  circuit.  As  is  seen,  iQ  is  a  unidirectional  current 


l.U 

( 

•^ 

Q 

^ 

x 

^^, 

__,, 

---' 

~~» 

^ 

'o 

5 

K. 

M. 

i1. 

* 

^^ 

^' 

f 

\ 

0 

/ 

\ 

/ 

^ 

/ 

( 

1 

1 

\ 

i 

* 

11  a 

1  i 

\ 

-•* 

-~- 

»»A 

i^ 

-^ 

<"4 

) 

Q 

1 

) 

11; 

^ 

?, 

() 

^- 

^ 

i() 

4 

0 

g 

(1- 

-  —  , 

ft 

) 

_^- 

1 

-.1 

\ 

/ 

\ 

.2 

( 

—.3 

\ 

/ 

N 

J 

1  0 

N 

, 

/ 

^ 

/ 

\  ' 

t-  A 

g 

\ 

/ 

\ 

1 

7 

\ 

/ 

\ 

1 

g 

1.9 

E 

M 

F, 

5 

1  8 

/" 

> 

\ 

4 

1  7 

\ 

2 

V 

lr 

/ 

/ 

\ 

\ 

/ 

^ 

1  fi 

t 

\ 

c 

/ 

\ 

•/' 

2 

/ 

\ 

/ 

\ 

\ 

/ 

/ 

\ 

\ 

/ 

1  4 

\ 

\ 

/ 

/ 

V 

, 

72 

0 

1  3 

91 

;/ 

li 

9 

2 

70 

p 

30 

4 

0 

/ 

b 

ip 

\ 

Gl 

0 

/ 

1   9 

V 

\ 

1 

\ 

N 

1 

/ 

o 

1    1 

A 

\ 

\ 

/ 

q 

1  0 

q 

N 

I 

7 

A 

/ 

—  4 

9 

/ 

A 

i 

^ 

-5 

/ 

\ 

1 

\ 

\ 

—  6 

7 

/ 

\ 

/ 

\ 

\ 

7 

g 

/ 

\ 

/ 

\ 

1 

\ 

5 

J 

^i 

K 

\ 

I 

._ 

/ 

9 

4 

N 

\ 

/ 

y 

/ 

1  0 

3  . 

\ 

\ 

;'  r 

/ 

/ 

\ 

\ 

/ 

\ 

I 

\ 

/ 

N 

\ 

/y 

\ 

i 

/ 

12 

) 

9 

0 

1 

^p 

2 

70 

\ 

3 

30 

/ 

4 

oO 

5 

IP 

6 

0 

\ 

I 

2 

V 

f 

\ 

\ 

i 

Q 

x^ 

^} 

\ 

4 

\ 

/ 

\, 

/ 

\ 

, 

^^ 

e 

-.6 

FIG.  38. 

of  slight  pulsation,  slowly  decreasing  in  magnitude.  After  a 
number  of  cycles  it  would  become  a  small  alternating  current,  as 
shown  in  the  curve  marked  Final  i0. 

This  feature  of  a  tuned  circuit  might  be  of  practical  importance 
in  connection  with  problems  of  rectification — charging  of  storage 
batteries  from  an  alternator  by  occasional  interruption  of  the 


CHARACTERISTICS  OF  CONDENSERS  81 

current  and  starting  it  at  the  time  when  normal  current  in  either 
branch  would  be  a  maximum. 

Circuits  of  Inductance  and  Capacity.  —  While  practically  such 
circuits  can  never  exist  they  offer  much  interest  from  a  theoretical 
point   of   view   since    their    study 
represents  an  introduction  to  oscil- 
lating circuits,  which  are  of  much 
importance  in  electrical  engineering. 

In  Fig.  39  is  shown  such  a  cir- 
cuit. In  practice  the  condition 
there  indicated  is  approached  FIQ 

when  a  very  low  resistance  over- 

head transmission  line  supplies  power  to  a  cable  net  work,  which 
case,  however,  is  fully  treated  in  a  subsequent  chapter. 

The  following  relation  exists  at  any  time. 


where  E  is  the  instantaneous  value  of  the  impressed  voltage  and 
e\  the  voltage  across  the  condenser. 

But  .  _  ndei 

-  dt 

thus  di  _       d*ei 

= 


dl* 
thus  LC^  +  ei  =  E 

or  d*ei    ,   _fi_       J^ 

dt2  ~^  CL~  CL 

The  solution  of  this  equation  has  been  given,  it  is 

ei  =  E  +  Aiemi' 
mi  and  ra2  are  the  roots  of  equation; 


m.  -  f  i  .  , 

<CL 


m 


82  ELECTRICAL  ENGINEERING 


But  e}'at  =  cos  at  +  j  sin  at 

€-jat  =  cos  at  —  j  s[u  at 

.'.A**  '   ™* 

+  j  sin  at(Ai  —  A2) 

=  Ao  cos  aZ  -f-  Bo  sin  at  =  A  sin  (a£  + 


where  A  and  5  are  integration  constants. 
The  current 


The  integration  constants  are  determined  from  the  knowledge 
of  the  initial  conditions. 

Assume  for  instance  that  it  is  desired  to  know  the  current  and 
the  voltage  across  the  condenser  at  any  instant  after  the  switch 
is  closed: 

At  the  time  t  =  0,  i  =  0  and  e\  =  0; 

.'.  0  =  E  +  A  smB 

and  1C         D 

0  =  A  -J    cos  B. 


Thus  cos  B  must  be:    0 

.'.  B  =      thus  sin  B  =  1  and  A  =  -  E 


and 


The  voltage  across  the  reactance  is  L  -j-  it  is 


=  cos 


CHARACTERISTICS  OF  CONDENSERS 


83 


In  Fig.  40  are  shown  the  voltages  across  the  condenser  and 
inductance  as  a  function  of  time.  It  is  seen  that  there  is  no  time 
at  which  the  voltage  across  the  inductance  is  greater  than  the 

impressed  voltage  and  it  is  also  seen 
that  at  the  instant  of  the  first  half 
period  the  voltage  across  the  con- 
denser is  twice  as  great  as  the  im- 
pressed. Thus  the  condenser  will  be 
subjected  to  double  voltage  during 
each  cycle. 

The  maximum  value  of  the  current 
is,  as  seen; 
FIG.  40. 


•>  f> 

/ 

-'" 

'*•• 

V1 

1.2 
.8 
.4 

/ 

7-) 

\ 

SI 

7 

v 

/ 

s 

\ 

/ 

\ 

. 

-.4 

\ 

/ 

s 

i 

-» 

-1.2 

^  — 

^**s 

<  ^T=2if\TcI.  Seconds  > 

The  frequency  of  the  alternating  current  /  =  - — ~/TT  *s  c 

the  natural  frequency.1     It  is  almost  always  much  higher  than 
the  frequencies  used  in  commercial  alternating-current  circuits. 

The  Discharge  of  a  Condenser  through  an  Inductance. — Re- 
ferring to  Fig.  41.  Let  E0  be  the  voltage  of  the  condenser  before 
the  circuit  is  closed  and  i  the  current  at  any  instant  after  the 
switch  is  closed. 


Then 


But  the  discharge  current 


e0  =  L 


thus 


and 


or 


i  =  -  C 
di 


di 
dt 

de( 
dt 


FIG.  41. 


L  dt  CL  dt2 

eo=  -CL'1 


di2 


CL 


the  auxiliary  equation  becomes : 

m2  +  -==•  =0   or  m  = 


1  In  case  of  a  transmission  line  the  natural  frequency  will  be  shown  to 
1 


84 
thus 

and 


ELECTRICAL  ENGINEERING 


cos 


—  cos 


The  integration  constants  A  and  B  are  determined  from  the  fact 
that  at  t  =  0,  i  =  0,  and  e0  =  E0.     Thus: 

EQ  =  A  sin  B 


0  =  - 


C        D 
—  cos  B 


'.  cos  B  =  0  and  B  —  ^ 


and 


.'.  sin  B  —    1  and  A 

t 

=  En  sm 


cos 


2 

E0 

—  I    =   En  COS 


sn 


s 


It  is  seen  that  the  discharge  frequency  is  the  same  as  the 
frequency  at  charge,  and  that  the  maximum  value  of  the  current 

P  ft 

E°\L- 

As  another  application  of  this  will  be  considered  the  condition 

when  a  short-circuit  is  suddenly 
opened  and  the  large  current  in- 
stantly interrupted. 

This  condition  is  diagrammat- 
ically  illustrated  in  Fig.  42.  S 
represents  a  switch  which  short- 
circuits  the  condenser  and  is 
opened  at  the  instant  under  consideration.  (In  practice  this 
switch  may  represent  a  short-circuit  across  the  cables  opened  by 
the  magnetic  effects  of  the  current.)  The  current  I  in  the 
short-circuit  is  evidently  the  same  as  the  current  in  the  inductance ; 
therefore  the  energy  stored  in  the  magnetic  field  is  0.5LJ2. 

At  any  time  after  the  switch  is  opened  the  current  i  flowing 
through  the  condenser,  inductance  and  generator  (all  assumed  as 


FIG.  42. 


CHARACTERISTICS  OF  CONDENSERS  85 

having  zero  resistance)  is  governed  by  the  condition  that  the 
energy  stored  in  the  condenser  and  inductance  is  the  same  as 
the  original  energy. 

.'.     0.5Li2  +  0.5Ce2  =  0.5L/2 

LI'2  -  Li*  =  Ce* 
but  .  _  ~de 

1      ^7. 

dt 
thus  „  /de 


,-2  =  C2/^V 

\dt) 
:.LP  -  LC*(~}2  =  Ce\ 


Differentiating  or/^2  ^2g  — 

~2LL    dt2   dt  = 

.    d*e        I 
-+6'- 


and  .     /     t  % 

e  '-'-  A  sin         +  B) 


for  t  =  Q,i  =  I,e  =  0 

Thus  0  =  A  sin  5;  and  sin  5  =  0,  and  B  =  0 

1C 

= 


i  =  I  cos 


It  is  interesting  to  note  that  while  at  the  instant  of  opening  the 
switch,  or  the  short-circuit,  the  voltage  e  across  the  condenser  is 
zero,  one-quarter  of  a  period  later  (period  being  here  the  natural 
period  which  is  extremely  short)  the  voltage  is  a  maximum  and 
is 

6max    = 

These  equations  are  instructive  in  that  they  show  that  the 
maximum  voltage  obtained  in  opening  a  short-circuit  in  a  cable 


86 


ELECTRICAL  ENGINEERING 


or  transmission  line  is  independent  of  the  length  of  the  line  and 
depends  only  upon  the  constants  of  the  circuit  per  unit  length  and 
the  current  at  the  time  the  circuit  is  interrupted. 

They  also  show  that  when  the  circuit  is  closed  on  a  transmission 
line  of  considerable  inductance  and  capacity,  the  maximum  rush 
of  current  is  also  independent  of  the  length  of  the  line  and  depends 
only  upon  the  value  of  the  e.m.f.  and  the  circuit  constants. 

Harmonic  E.m.f.  Impressed  upon  a  Circuit  of  Inductance  and 
Capacity  but  Negligible  Resistance. — This  strictly  theoretical 
condition  is  chosen  for  two  reasons.  The  solution  of  the  equa- 
tions introduces  some  mathematical  operations  which  have 
hitherto  not  been  considered  and  the  problem  from  the  electrical 
point  of  view  illuminates  in  a  relatively  simple  way  what  happens 
in  the  extreme  case  in  switching  high-potential  circuits. 

The  general  equation  obviously  becomes: 


E  sin 


ei 


(1) 


where  e\  is  the  voltage  across  the  condenser,  Fig.  43. 
But 


^  dt 

.'.  E  sin 


thus  L  -7 
at 


CL 


or 


FIG.  43. 

d26i  61  E 

W  +  CL  =  CL 


sln 


(2) 


It  is  seen  that  the  right-hand  member  of  the  equation  is  a 
function  of  L  To  solve  such  equation  the  solution  of  the  comple- 
mentary function  is  first  found;  that  is,  zero  is  substituted  for 
the  right-hand  member: 


V   'r  CL  ~ 

i2  +-==  =  0     or  m  =  ±  j  .J— 

CL/ 


(3) 


CHARACTERISTICS  OF  CONDENSERS  87 

Ai€+A/cE'    +  AvT^m,'    =  A^at  +  A#-iat        (4) 


where 


The  equation,  as  has  been  shown  previously,  can  be  written 

61  =  A  sin  (at  +  #) 
Thus  the  complementary  function  is 

eQ  =  A  sin  (at  +  B)  (5) 

The  next  step  is  to  eliminate  the  sine  function  from  the  general 
equation  (2)  by  two  successive  differentiations: 


Substituting  the  value  of  Ea2  sin  coi  from  equation   (2)  and 
arranging  the  equation  in  the  order  of  the  derivatives: 

(7) 


The  complete  solution  of  (7)  is  obtained  in  the  usual  way: 
m4  +  (a2  +  a>2)ra2  +  coV  =  0 
W2(m2  +  a2)  +  a>2(ra2  +  «2)  =  0 
or  (m2  +  w2)   (m2  +  a2)  =  0 

.*.  m  =  +  jb) 
m  =  +  jo. 
Thus  d  =  Ai  sin  (at  +  #0  +  A2  sin  (a«  +  52)  (8) 

By  referring  to  (5)  it  is  evident  that  A2  and  52  must  be  the  same 

as  A  and  B. 

Thus  €i  =   eQ  +  AI  sin  (o>£  +  BI) 

The  integration  constants  AI  and  BI  are  determined  from  the 
fact  that  the  expression  A\  sin  (co£  +  BI)  must  be  a  particular 
solution  of  (2). 

Thus  «i  =  AI  sin  (o>£  +  £1) 

-7-  =  AICO  cos  (&t  -}-  BI) 

dzd 

-       =  --  Aiw2  sin  (cot  +  BI) 


88  ELECTRICAL  ENGINEERING 

Substituting  these  values  in  (2). 
Thus 

-  Aico2  sin  (ut  +  Bi)  +  Ai«2  sin  (coZ  +  BJ  =  Ea*  sin  ut     (9) 

or  A^a2  -  co2)  sin  (at  +  BI)  =  Ea2  sin  ut. 

Thus  equating  the  coefficients  of  similar  terms: 
Ea2  =  Ai(a2  -  co2) 

* --^.-5%;  (10) 

or  —  or      £c  —  a: 

and  £1  =  0  /.  from  (8)  (11) 

ei  =     EXc     sin  ut  +  A  sin  (erf  +  5)  (12) 

3?c         32 

The  second  term  in  this  equation  may  in  this  case  be  more 
advantageously  written : 

AQ  sin  at  +  Bo  cos  a£ 
/.  ei  =  -       -  sin  coi  +  A0  sin  a^  +  Bo  cos  at  (13) 

3?c         «E 

=  7a:c  sin  coi  +  A0  sin  a^  +  Bo  cos  at  (14) 

Where  /  is  the  maximum  value  of  the  permanent  current, 
that  is, 

/  -  ~^—  (15) 

Xc   -  X 

i  =  C  3  -  =  Ixcu  cos  ut  +  AQa  cos  at  —  B0a  sin  CK£        (16) 
at 

Considering  the  problem  of  starting  a  current  in  such  a  circuit : 
when  t  =  ti,  i  =  0  and  e\  =  0.  If  these  values  are  substituted 
in  equations  (13)  and  (16)  it  is  readily  found  that 

BQ  =  Ixc     —  sin  ati  cos  co£i  —  sin  at\  cos  cd\  (17) 

AQ  =  —  Ixc     sin  ati  sin  wt\  -\ cos  at\  cos  co^i  \         (18) 

Substituting  these  values  in  the  equations  for  the  current  and 
e.m.f.  across  the  condenser  we  find: 

ei  =  -  sin  at cos  coZi  sin  a(t  —  ti) 

Xc       X  L  a 

-  sin  coii  cos  a(t  -  fi)l        (19) 


CHARACTERISTICS  OF  CONDENSERS  89 

and 

i  =  C  -TT   =  ~  cos  at  —  cos  coti  cos  a(t  —  ti) 

at        Xc  —  x  L 

1C  1 

+  xc  ^Ij-  sin  coti  sin  a(t  -  fc)  I        (20) 

While  as  a  rule  equations  of  the  form  given  in  (2)  having  a 
sine  function  or  exponential  function  on  the  right-hand  member, 
can  be  solved  by  the  method  given  above,  it  may  be  opportune 
here  to  call  attention  to  another  well-known  method  of  more 
general  application. 

The  differential  equation  is: 

fJZf, 

<L     +  eia2  =  Ea2sin  cot 


This  expression  is  given  in  symbolic  factors  as  follows: 

(D  -  mi)  (D  -  m2)ei  =  Ea2  sin  cot  (21) 

where  D  is  an  abbreviation  of  -r  and  mi  and  mz  are  the  roots  of 
the  complementary  function 

d2ei 

777-  -f-  eia2       or     m2  +  a2  =  0 
at 

.'.  m  =  ±  ja  or     mi  =  +  ja 
and  mz  =  —  ja 

Equation  21  can  thus  be  written 

(D  +  ja)  (D  -  ja)ei  =  Ea2  sin  cot 
Let  u  =  ei(D  —  ja) 

.'.   (D  -f  ja)u  =  Ea2  sin  cot 

^  +  jau  =  Ea2  sin  coi  (22) 

This  is  a  linear  differential  equation  of  the  first  order  and  its 
solution  is. 

u  =  e-i*t  |  6+/««  j^o:2  sin  w^  ^  +  Ci€~^  (23) 

/.  .    .        .   ,.          .  ,  /  ja  sin  cot  —  co  cos  coA 
eiat  sm  cot  dt  =  e]at  [  J— 
\  co2  -  a2  / 

cot-^co^cot) 

« 


90  ELECTRICAL  ENGINEERING 

Since  u  =  e^(D  —  ja) 


dt 

i  =  e+3'at      e-*atudt 


I  e- 

e]'at  I  €~}'at  •  -  -  —  „  (ja  sin  cot  —  co  cos 
J  to2  —  a2 

t    |    e- 


+   €iat         e-jat    ClC-J««  dt   +    C  #>*'  (25) 

These  four  integrals  are  solved  independently  below  : 
First, 


s«3  .*f. 

-«2'  Jf 


sn 


(          —  la  sm  co*  —  co  cos  co  *\                     EC 
,— jat          J I     —    A  
co°  +  a2               /              J  co4+a2 

(ja  sin  co*  +  co  cos  co*)      (26) 
Second, 

r 

~iat  cos  co^  dt  = 


—  a*          I  co"  —  a* 


2  - 


I   -jat  u  s^n  ^  ~  Ja  cos  ^  \  •    ^a?co 

\  C"  co2  +  a2  /   :      ~  J  co4 -a4 

(a  cos  co*  —  jco  sin  co*)      (27) 


Third, 

r     _.at    _     _.at  r  _2.a< 

O  1C  I      C  €  (Zt'      vy  l6  I      C  ttt'      

J  J 

"    2jae'a<  €~23at  =  2jae~}'at    =  C^at        (28) 

Fourth,  C2e>'at  (29) 

The  last  two  terms  can  be  written 

C3e-'a<  +  C2e>°<  =  C4  sin  (a*  +  C6)  (30) 

In  the  general  equation  it  is  seen  that  the  second  integral  is 
negative  thus: 


co2  sin  co£  +  a2  sin  co^  —  jaco  cos  co^         (31) 


=  C4  sin  (at  +  C6)  H  --  ri    jaw  cos 

+  co2  si 

n      -     /    * 

=  C4  sin  (at 

sin  (« 
This  equation  is  identical  with  (12). 


=  C4  sin  (««  +  CB)  +  --  sin  co*  (32) 

Xc         X 


CHARACTERISTICS  OF  CONDENSERS  91 

As  an  application  of  these  formulae  will  be  considered  a  100- 
mile  60-cycle  transmission  line  supplying  power  to  a  cable  net- 
work of  50  miles.  For  the  sake  of  simplicity  and  for  the  sake  of 
later  instructive  comparison  the  resistance  of  the  cable  and  of 
the  overhead  line  will  be  neglected  in  this  particular  investigation 
and  it  will  also  be  assumed  that  the  inductance  of  the  cables  and 
the  capacity  of  the  overhead  lines  are  so  small  as  to  be  negligible 
when  compared  with  the  inductance  of  the  overhead  line  and  the 
capacity  of  the  cables. 

While  the  inductance  of  a  line,  of  course,  depends  upon  the 
size  of  the  conductors  and  the  distance  between  them,  in  reality 
it  is  not  subject  to  a  great  deal  of  variation  in  ordinary  lines. 
It  is  about  0.002  henrys  per  mile  of  single  conductor. 

The  capacity  of  a  cable  system  is,  however,  subject  to  great 
variation,  depending  upon  the  nature  of  the  cables.  Assume 
that  in  this  case  it  is  2  m-f.  per  mile  of  single  conductor,  when 
referring  to  neutral  voltage: 

C  =       4  farads  and  L  =  0.2  henrys 


=  26.4  ohms 

x  =  2irfL  =  75  .4  ohms 
„  =  27T/  =  377. 

If  the  circuit  is  closed  when  the  impressed  e.m.f.  is  zero,  that 
is,  when  ti  =  0,  then  equations  209  and  210  become: 

ei  =  -  0.54#[sin  377*  -  1.69  sin  223*] 
and  i  =  -  0.0204#[cos  377Z  -  cos  223*] 

The  time  for  one  complete  cycle  of  the  fundamental  wave  is 
gg  =  0.0166  sec. 

If,  therefore,  the  circuit  is  closed  when  the  impressed  e.m.f. 
is  a  maximum,  that  is,  when 

h  =  t  =  0.00416 
then  the  equations  become: 

d  =  -  0.54#[sin  377*  -  cos  223(*  -  0.00416)] 
i  =  -  0.0204#[cos  377*  +  0.59  sin  223(£  -  0.00416)] 


92 


ELECTRICAL  ENGINEERING 


These  curves  are  shown  in  Figs.  44  and  45  when  the  impressed 
e.m.f.  is  100  volts. 

The  curve  e\  in  Fig.  44  shows  the  e.m.f.  across  the  condenser, 
the  curve  i  the  current  when  the  switch  is  closed  at  zero  value  of 


4 

^-~, 

/ 

\, 

/ 

\ 

2100 

^_ 

/ 

\ 

/ 

N. 

^ 

•> 

/ 

^^ 

sci 

Y 

"X 

' 

V 

, 

m 

/ 
/ 

/ 

V 

? 

V 

\ 

/ 

/ 

\ 

V 

/ 

\ 

s 

N 

\ 

55  1    50 

/ 

/ 

/ 

\ 

\ 

\ 

v/ 
cY 

/ 

\ 

/ 

/ 

\ 

\ 

^0  *0 

/ 

.X 

\ 

\ 

\ 

0 

/ 

\ 

/ 

\ 

/ 

"-—  . 

—  \ 

o      > 

£ 

0 

i 

^0 

\ 

2 

70 

\f 

30 

^ 

oO 

b 

ty 

6 

JO 

\ 

I 

0 

8 

LO 

/ 

yu 

a  1     50 

\ 

\ 

i 

A 

/ 

a. 

t 

A 

A 

/ 

\ 

\ 

$/ 

\ 

/ 

/ 

\ 

/ 

\ 

x 

i 

> 

v  \^ 

^ 

\ 

/ 

\ 

/ 

/ 

U  2    100 

\ 

V 

/ 

\ 

/ 

V. 

^x 

\ 

/ 

4 

s^ 

FIG.  44. 


2  200 

150 


.S  0  oO 
50 


1  100 
150 

2  200 


90 


isp 


270 


\ 


/450 


720 


9(0 


FIG.  45. 

the  impressed  e.m.f.  The  corresponding  lines  in  Fig.  45  show 
the  same  quantities  when  the  switch  is  closed  when  the  impressed 
e.m.f.  is  a  maximum.  In  both  figures  the  dotted  sine  wave  is 
the  impressed  e.m.f. 


CHARACTERISTICS  OF  CONDENSERS 


93 


Circuits  Having  Resistance,  Inductance  and  Capacity.  Con- 
stant Impressed  E.m.f.  —  Let  E,  Fig.  46,  be  the  constant  impressed 
e.m.f. 

r,  the  resistance 

L,  the  inductance 
and 

C,  the  capacity  in  farads. 


L 

c: 


Then 


FIG.  46. 

T  di    , 


Differentiating  di  d*i       i 

=  T+L         + 


or 


dH   >rdii     _ 
dt*  +  L  dt  +  CL  " 

The  solution  of  this  equation  has  been  shown  previously  to  be  : 


Where  mi  and  ra2  are  the  roots  of  the  auxiliary  equation. 


where 
and 


_ 
2L 


4L2       CL 
CL 


2L  ~         = 


=  - 


- 


4L 


94  ELECTRICAL  ENGINEERING 

Three  conditions  are  possible: 

(a)  r2  •     -T^  is  positive. 

(b)  r*  -    -~r  is  negative. 

0 

4L. 

(c)  r2  —  -fT  is  zero. 

Considering  first  the  Case  (a), 

^       ~^r  (positive). 

Then  i  =  ^[A*?  +  A*-'*1}. 

The  e.m.f.  across  the  condenser  is: 

ei  =  E  -ir  -Ld~  =  E  -  re-at[A^1  +  A*'*1]  -  L[((3  -  a 


The  integration  constants  for  starting  the  current  in  this  circuit 
are  determined  from  the  fact  that  when  t  =  0,  i  =  0  and  e\  =  0. 

Thus  E 

Ai  =  —  Ai  and  AI  =    ^j— 
ZLp 


2L/3 


2       4L 

\r      '^:=>S> 


If 


then  .       El    -l'-=^)i        -f'J^\t\ 

*-•§(•  -•  )• 

By  differentiating  i  and  substituting  its  value  and  that  of  i  in 
the  equation  of  the  voltage  of  the  condenser: 

e\  =  E  —  ir—  Jl  -j-.j 
dt 

we  get 

[1    /  r~st  r+st\~] 

i          •    if«._i_  cr\  — o7~ i        (^       Q\f — 9r~t\ 

1  —  ^.^  I  (r  •+-  o)€     2L  —  IT  —  ^Jc      *^ 
^o  \  /  J 

The  equation  for  the  discharge  current  can  readily  be  proven 
to  be  exactly  the  same  as  that  of  the  initial  current  except  for 
reversed  sign. 


CHARACTERISTICS  OF  CONDENSERS  95 

During  the  discharge  the  voltage  is 


r-f  S 


where  E0  is  the  voltage  of  the  condenser  before  the  discharge. 
Case  (6).-  r«  -  4  £  negative. 

In  this  case  +  r2 ~-  can  be  written  j  -\\-~  —  r2 

\  C  \  C 


Si  where  Sx  ==  -  r2 


It  has  been  shown  previously  that  this  equation  can  be  written  : 

(1) 


where  A  and  5  are  integration  constants.     The  e.m.f.  across  the 
condenser  is 


Substituting  the  values  of  i  and    -  as  obtained  from  equation  (1) 


Counting  time  from  the  instant  of  closing  the  circuit,  then  for 
t  =  0,  i  =  0,  and  e\  =  0. 

From  equation  (1)  it  is  found  that  A  sin  B  =  0;    .'.  B  =  0, 

2# 

and  from  equation  (2)  A  =  -~-  • 

01 

.          2E          *t      .      Sit 

•••t  =  ST6    2L    Sm2L 


where  tan  7  =  —  • 

In  a  similar  way  is  found  the  equation  for  the  discharge 
current  which  will  be  identical  with  the  charging  current,  and 

the  voltage  across  the  condenser,  ea  =  EQe  ~2Ll  sin  (  ^F  +  T  )  > 
where  EQ  is  the  original  voltage  across  the  condenser. 


96  ELECTRICAL  ENGINEERING 

Case  (c). — 


4L 

r2 


In  this  case,  as  has  been  shown  previously, 

i  =  emi*  (A  +  Bx) 

r 

i  =  e    2L   (A  +  Bt) 


or 


=  E 


re~2L  l  (A  +  Bt)  -  LBe    2Ll   +  ^~- 


If  the  time  is  counted  from  the  instant  of  closing  the  switch, 
then  for  t  =  0,  i  =  0,  and  ei  =  0 

A  =  0 
and  -       "       -  —         ^      hi 


0 


E  -  LB,  or  B 


and 


The  equation  for  the  discharge  current  is  the  same  as  that  of 
the  initial  charging  current,  the  voltage  across  the  condenser 
during  discharge  being: 


As  an  application  of  these  equations 
will  be  considered  the  case  of  starting 
a  direct  current  at  500  volts  in  a  20-mile 
concentric  cable  having  the  dimensions 
given  in  Fig.  47. 

In  this  instance  it  will  be  assumed 
that  the  capacity  of  the  line  can  be 
represented  as  that  of  a  condenser  at 
the  end  of  the  line  taking  one-half  of 
the  charging  current. 

It  will  be  assumed  that  the  specific 
inductive  capacity  is  3,  the  diameter 
of  the  inside  conductor  0.5  in.,  the 
inside  diameter  of  the  outside  conductor  0.7  in.,  and  the'  out- 
side diameter  0.86  in.  The  resistance  of  the  20  miles  of  cable  is 
8.8  ohms. 


FIG.  47. 


/ 

CHARACTERISTICS  OF  CONDENSERS  97 

The  capacity  per  mile  of  concentric  cable  previously  given  is 

n  o^sft  Jf 
Cmf  =  -      — ~—  =  0.795  m-f.  per  mile. 

^og- 

Thus  the  capacity  of  the  20  miles  of  cable  is  15.9  m-f.  and  the 
equivalent  capacity  at  the  end  of  the  line  is  7.95  m-f.,  or  7.95 
by  10~6  farads. 

Since  the  determination  of  the  inductance  of  a  concentric 
cable  involves  the  general  method  applied  to  other  systems,  it 
will  be  given  below,  although  such  determinations  do  not  come 
within  the  scope  of  this  treatise. 

The  inductance  is  recollected  to  be  numerically  equal  to  the 
interlinkages  of  the  turns  and  flux  per  unit  current. 

In  general  if  the  m.m.f.  acting  in  a  circuit  is  M  then  the  flux 

4irM  X  area  of  magnetic  circuit 

produced  is  — , -r — * —        ~r- = ^ The  interlinkage 

length  of  magnetic  circuit 

factor  is  that  fraction  of  the  total  current  which  is  enclosed  by 
the  flux  and 

L  =  y-2J  flux  X  turns  X  interlinkage  factor. 

Consider  first  the  flux  in  the  inside  conductor  due  to  the  assumed 
uniform  distribution  of  the  current. 

At  a  distance  x  from  the  center  see  Fig.  47,  the  m.m.f.  is  — ^  / 

where  /  is  the  total  current.  The  area  of  the  flux  per  centimeter 
of  length  of  conductor  is  dx  and  the  length  of  the  magnetic  circuit 
is  2irx. 

x2     dx  x 

...dv,1  =  4,_/_==2J_&   ( 

TTX2 

This  flux  interlinks  with  — ^  of  the  total  current,  and  hence  the 

x2 
interlinkage  factor  is  —^. 

1   Cr     x3 

.'.  LI  =       I    27—  dx  =  %  (assuming  /x  =  1) 

Between  the  conductors  the  flux  interlinks  with  the  whole 
current,  and  hence  by  a  similar  reasoning  we  get 


98  ELECTRICAL  ENGINEERING 

The  current  in  the  inner  conductor  interlinks  with  the  entire 
flux  which  is  inside  of  the  outer  conductor  but  which  is  caused  by 
the  difference  in  m.m.f  .  in  the  inner  and  outer  conductor. 
At  a  distance  XQ  the  m.m.f.  is  thus 


_ 


R02-R2*       *RQ2-R2 

The  interlinkage  of  this  flux  with  the  current  in  the  inner  con- 
ductor is  of  course  unity;  therefore 

L*  =  I J  ~*o  flo2^!^  dx°  =  R»2  -  R2  log  ^  " "  l 
The  inductance  of  the  outer  conductor  should  be  added  to  give 
the  total  inductance  of  the  cable. 

The  m.m.f.  is  _  r  RQ    ~  XQ 


Ro2-  1 
The  interlinkage  factor  is 

xp2  -  R2 
-R2 

,2    _ 


•'•  L*  -  -  7/ 


z0  (Ro2-R2) 

^  RQ2  -  R2  +  (R02-R2)2  log  ~R 

The  total  inductance  L  =  LI  -j-  L%  +  Ls  +  LI  which  is  readily 
proven  to  be: 

This  inductance  is  expressed  in  the  absolute  system  of  units. 
By  dividing  by  109  the  inductance  is  expressed  in  henrys. 

The  combined  inductance  L  =  0.0039  henrys;  thus  r  =  8.8, 
L  =  0.0039,  and  C  =  7.95  m.f. 

r2 =  77.5  —  1960  is  thus  negative. 

c 

Therefore  this   problem   comes  under  the   second   case   and 


t  =  ^/%-  -  r2  =  43.4. 
^  =  5550  ~  =  1130  tan  7  =  — -  =  4.93r  =  78.5 

LU  ZLi  T 

=  1.37  radians 
/.  i  =  23e-1130'  sin  5550* 
and  ec  =  500[1  -  1.02e-1130(  sin  (5550*  +  1.37)]. 


CHARACTERISTICS  OF  CONDENSERS 


99 


The  frequency  of  the  oscillation  is 


5550 

27T 


=  885  cycles,  and  the 


time  for  one  oscillation  0.00113  sec. 

The  maximum  value  of  the  current  is  determined  by  differen- 
tiation. It  occurs  when  t  =  0.000246  sec.  When  5550^  = 
78.5°,  the  current  is  17.1  amp.  The  next  maximum  value  occurs 
when  t  =  0.000246  +  0.00113  =  0.001376  sec. 

The  maximum  value  of  the  voltage  across  the  condenser  is  also 
determined  by  differentiation.  It  occurs  when  t  =  0.000565 
sec.,  when  ec  =  763  volts.  The  next  high  value  occurs  obviously 
at  t  =  0.001695  sec. 

These  curves  are  shown  in  Fig.  48. 


0004 


Starting 


V;  Condenser  Volt;  K 


urrent 


m- 


.001 


odd! 


0012 


0014 


0016 


.0018 


400 


200 


,002 


FIG.  48. 

It  is  of  interest  to  note  that  for  a  given  distance  of  transmission 
the  capacity,  and  therefore  the  charging  current,  is  several  times 
as  great  in  the  case  of  the  concentric  cable  as  in  the  case  of  the 
cable  with  parallel  wires. 

Similarly  the  inductance  is  several  times  as  great  in  the  case  of 
an  overhead  line  as  in  the  case  of  the  cable.  As  a  second  numer- 
ical application  of  these  equations  will  be  considered:  100  miles 
of  overhead  transmission  line  supplying  energy  to  a  cable  network 
50  miles  in  length. 

It  will  be  assumed  that  the  cable  system  consists  of  a  large 
number  of  short  cables  projecting  in  different  directions  from  the 
terminal  substation,  as  would  be  the  case  when  a  high-tension 

7 


100  ELECTRICAL  ENGINEERING 

line  supplies  energy  to  a  city  lighting  network.  The  resistance 
of  the  cable  system  can  therefore  be  neglected.  It  will  be 
assumed  that  the  high-potential  line  is  three-phase  and  consists 
of  No.  00  B.  &  S.  wire,  having  a  resistance  per  100  miles  of  40 
ohms  and  an  inductance  of  approximately  0.2  henry.  Hence  the 
capacity  of  the  overhead  line  is  very  small  compared  with  that  of 
the  cable  and  it  will  be  neglected. 

The  problem  is  to  determine  the  values  of  the  current  and  vol- 
tage across  the  condenser  when  a  steady  e.m.f.  of  E  volts  is 
applied  at  the  generating  station. 

E  =  100 

r  =  40 

L  =  0.2 

C  =  0.0001  farad. 
/.  r2  =  1600 

-Q    =  0.8  X  104  =  8000. 

4L. 

.'.  r2  =  -^  is  negative. 

Therefore  there  is  an  oscillation  when  the  switch  is  closed,  and 
the  constants  are  to  be  obtained  from  case  (6). 


=  2=  80'  7  =  !o  =  2"  tan^  =  2 

and  r         40  Si        80 

V  -  63.5  2L  =    ^4  =:  10°>  2L  =  04  = 

:.i  =  0.025#e-  10°*  sin  160* 
and  a  =  E[l  -  e"100'  1.12  sin  (160*  +  63.5°)]. 

The  time  for  a  complete  cycle  is  T^T  =  0.0392  sec.,  corresponding 

to  a  natural  frequency  of  ^  =  25.5  cycles  per  sec.     It  is  inter- 

esting to  see  that  the  effect  of  the  resistance  is  to  lower  the 
natural  frequency,  since  if  the  resistance  is  neglected  it  would  be 

—7=  =  35.5  cycles. 


CHARACTERISTICS  OF  CONDENSERS  101 

Circuit  Containing  Resistance,  Inductance  and  Capacity  in 
Series.  Harmonic  E.m.f.  Impressed. — Fig.  49.  From  previous 
discussions  it  is  evident  that  the  general  equation  is 

di 

dt 


E  sin  cot  =  ir  -\-  L  ^-  -f-  ~-  I  idt  (1) 


°r  E  sin  0  =  ir  +  x~  +  xc  (idd  (2) 


FIG.  49. 

The  latter  form  is  preferable  when  dealing  with  alternating- 
current  phenomena,  but  of  course  it  must  be  remembered  that  x 
and  xc  refer  to  the  impressed  frequency  and  not  to  the  natural 
frequency  of  the  system;  that  is, 

x  =  2-nfL  and  xc  = 


The  solution  of  (2)   can  best  be  obtained  by  differentiating 
twice, 

di  d2i 

E  cos  6  =  r  -TT  +  x  -7T-2  +  xci  —  E  sin  6 


cttf 

dzi  d*i  di 


dl  '  idB  (3) 


Differentiating  (3)  and  rearranging  the  equation: 
d*i          dH  d*i          di] 


Xcl  =  ° 


The  auxiliary  equation  is: 

xm*  +  rra3  +  (xc  +  x)m2  +  rm  +  xc  =  0 
ra2(zra2  +  rm  —  xc)  +  xm2  +  rm  +  xc  =  0 
/.  (m2  +  1)     (xm2  +  rm  +  xc)  =  0 

Vr2  - 


Let  r 

2x 


102  ELECTRICAL  ENGINEERING 

then  mi  =  +  j 

W2  =  —  j 

m3  =  —  a  +  ft 
1714  =  --  a  —  ft 

and          i  =  A,  sin  (0  +  A2)  +  ^'(A*-*8  +  A,^e)  (6) 

The  integration  constants  A  \  and  A  2  could  be  found  by  methods 
outlined  in  the  chapter  on  circuits  of  inductance  and  capacity. 
It  is  possible,  however,  for  students  familiar  with  elementary 
electrical  engineering  to  determine  them  at  once. 

Apparently  the  first  term  represents  the  permanent  and  the 
second  the  transient  condition.  In  permanent  operation  the 
current  leads  or  lags  behind  the  impressed  e.m.f.  by  an  angle  </> 
which  depends  upon  the  numerical  values  of  the  two  reactances. 
The  final  value  of  the  current  becomes,  then, 

-T7 

i  =  ^  sin  (0  +  0) 

^o 

if  tan  4  =  —^  and  Z0  ==  vV  +  (xc  -  x)  2  (7) 

.'.  i  =  ~  sin  (6  +  </>)+  <raV3/'  +  A#-ft°)  (8) 

&Q 

The  integration  constants  A3  and  A  4  depend  upon  the  terminal 
conditions. 

Before  proceeding  farther  it  is  well  to  discuss  the  possible  con- 
ditions, namely: 

(a)  r2  •-  4xxc  is  positive. 

(6)   r    •-  4xxc  is  negative. 

(c)    r    •-  4xxc  is  zero. 
Case  (a).— 

r2  —  4xxc  is  positive. 

Here  ft  is  real  and  the  solution  of  i  is  given  in  equation  (8). 
Case  (&).— 

r2  —  4zzc  is  negative. 

In  this  case  \/V2  —  ±xxc  =  j  V4xxc  —  r2  (9) 

Let 


then  i  =  lj-  sin  (6  +  <l>)  +  AiT'to  sin  (0J  +  j)  (11) 

^0 

where 


CHARACTERISTICS  OF  CONDENSERS  103 

and  ,  AB  As  +  A4 

7  =  tan"1  -r-  =  tan"1  7-1— 

A6  (A3  -  A4) 

a  S°       t  =  ~-  sin  (0  +  </>)  +  €~"*[A6  cos  0i0  +  A6  sin  faO]      (12) 


(c).—  r*  - 

If  (12)  is  true,  then  m3  =  w4,  and  we  do  not  have  a  general 
solution;  that  is: 

-  a  +  0  =  -  «  -  0  since  0  =  0  (13) 

A  general  solution  is  obtained  by  letting 

w4  =  ra3  +  h  (14) 

where  h  is  very  small. 

Here  w3  =  —  a  and  ra4  =  —  a  +  h 

then  (see  also  equation  42,  Chap.  II), 

E 

i  =  -yr  sin  (0  +  0)  +  A3e~'  '  -f  A4e(~  (15) 

which  may  be  written: 

i  =  £   sin  (0  +  0)  +  e~ae[A8  +  A90]  (16) 

where  A8  =  A3  +  A4  and  A9  =  A4/i. 

Each  case  will  be  considered  independently. 
Case  (a). —  r2  —  4xxc  (positive). 

Since  r2  —  4xxc  is  positive,  0  is  a  real  number  and  the  solution 
of  the  differential  equation  (4)  is : 

i  =  ^-  sin  (0  +  0)  +  A3e~(  *   ®6  -\~  A4e~(  (17) 

Let  («  -  0)  =  K  and  (a  +  0)  =  K±  (18) 

By    differentiating    equation    (17)    and    substituting    in    the 
equation 

d  =  E  sin  0  —  ir  —  X-J-Q,  (19) 

the  voltage  across  the  condenser  is: 

1?7 
ei  =  E  sin  0  -  -^  sin  (0  +  0  +  ¥)  +  A3e~J 

^lfl(^ia:  -  r)     (20) 
where  .,.      x  (21) 

(22) 


104  ELECTRICAL  ENGINEERING 

If  the  problem  is  to  find  the  current  and  the  condenser  voltage 
at  any  instant  after  the  circuit  is  closed,  and  if  the  circuit  is 
closed  when  0  =  0i,  then  i  =  0  and  ei  =  0. 

Substituting  these  conditions  in  equation  (23),  it  may  be 
written : 

Ase~Kei  +  A,*-*1'1  =  --  |^sin  (0!  +  0)  (24) 

Also  equation  (20)  can  be  written: 

EZ 
Ax'™1  (Kx  -  r)  +  A4e~J    9l  (Kix  -  r)  =  -^-  sin 

(0i  +  0  +  ^)  -  E  sin  0i     (25) 
Solving  equations  (24)  and  (25)  for  A3  and  A4  we  have: 

E 

A3  =  €K01  [Z  sin  (0i  4-  <f>  +  ^)  + 

(Kix  —  r)  sin  (0i  -f  0)  —  Z0  sin  0X]     (26) 
and, 

E 

— ^r  eKl01  \Z  sin  (0i  +  d>  +  ^)  + 


x  -  r)  sin  (0i  +  0)  -  Z0  sin  Oi]     (27) 


Case   (6).  —  As  stated  before    the    expression    \A*2  —  £xxc   is 
imaginary  and  from  equation 

(28) 


Equation  (11)  may  be  written: 

E7 

z  =  ^-  sin  (0  +  0)  +  e~a0[A5  cos  |8i0  +  A6  sin  j3i0]      (29) 

-^0 

where  A6  =  A3  +  A4  and  AG  =  j(A3  —  A4). 

If  the  switch  is  closed  at  0  =  0i,  e\  =  0  and  i  =  0.     From 
these  conditions 

A  5  =  [-  f-  sin  (0!  +  </>)  -  A6e-afll  sin  /3101]-^-     (30) 

L        Z/o  J  COS  Piwi 

From  these  relations  and  the  equation  (19) 

can  be  written 
- 
-  r)  cos  0i0]  +  A6[-  $ix  cos  0i0  +  (ax  -  r)  sin 


di 
e\  =  E  sm  0  —  zr  —  x  -> 


EZ 

=  E  sin  0  -  -=r-  sin  (0  +  0  +  ^)  +  e~a0[A5[/3iZ  sin 


CHARACTERISTICS  OF  CONDENSERS  105 

where 

Ecosp,d}   ae[Z0    . 
A6  =  -    TToT-     e         r  sm  0i  -  cos  (01  +  0)  -  01 


and  [#  sin  (^i  +  0)ea<?1  +  A6^o  sin  ffifln 

Z0  cos  /3i0 

Case  (c).— 

(r2  =  4a:xc),  or  the  "critical  case." 

Equation  (16)  may  be  written: 

E 

i  =  -y-  sin  (0  -f-  <p)  -{-  A%e    '     -f-  ^4.9^e    '  (31) 


If  as  before  th'e  switch  is  closed  when  0  =  61  then  i  =  0  and 
e\  —  0.     From  these  relations  and  the  equation  (19) 

di 

ei  =  E  sin  6  —  ^r  —  x  -^ 
uu 

the  condenser  voltage  is  found  to  be: 

777 

e,  =  E  sin  6  -   -^-  sin  (0  +  0  +  ¥)  +  e~a5[^8(o!X  -  r)  + 

^9(ax  -  a:  -  r)]      (32) 

where, 

77         r^  n 

A9  =  -=r  e"01"0  sin  (9T  -  cos  (Bl  +  <j>)  -  a  sin  (0i  +  </>)     (33) 
Z/o          L  ^ 

A8  =   -  ~  e^1  sin  (19!  +  0)  -  A90  (34) 


CHAPTER  V 

A  CIRCUIT  CONTAINING  DISTRIBUTED  RESISTANCE 
AND  INDUCTANCE 

An  aerial  transmission  line  with  negligible  capacity  and  leakage 
conductance  is  an  example  of  such  circuit. 

Fig.  50  represents  an  aerial  transmission  line  with  negligible 
capacity  and  leakage  conductance  and  with  a  load  having  an 
impedance  of  \/Ri2  +  Li2co2. 

Measuring  x  from  the  receiving  end,  consider  an  element  of 
the  line  dx. 


Let  the  resistance  of  the  line  be  R  ohms  per  unit  length  of  the 
conductor  and  the  inductance  L  henrys  per  unit  length. 

Then  the  resistance  of  the  element  is  Rdx  and  its  inductance 
Ldx,  and  the  voltage  across  the  element  is  de.  Therefore, 


or 


de  =  Rdxi 


de 


TJ      di 
MXdf 


di 


As  the  same  current  flows  in  all  parts  of  the  line  i  is  not  a 
function  of  x,  thus  equation  (1)  is  readily  integrated,  it  is: 


106 


RESISTANCE  AND  INDUCTANCE  107 

where  T    di 

x  =  0,  e  =  R]i  +  Li  -T  , 

.        =  di 

fn.   .    Tdi\  r  di 

.   .  e  =  (/fa  +  L--i-  iz  -f  72 1*  +  Lrr  (2) 

\  d<  /  d/ 

where  x  =  I,  e  =  E  or  E  sin  o>Z,  depending  upon  whether  the 
generator  voltage  is  constant  (a)  or  alternating  (6),  therefore, 


(a)      E  =     Ri  +  LJJ   I  +  R,i  +  Lr      =  (fl/  +  R,)i  + 

(JH+Zu)^    (3) 
(6)  £  sin  co*  =  (Rl  +  720*  +  (^  +  LI)^  (4) 

/??  +  Ri  is  the  total  resistance  and  LI  -f  ^i  the  total  inductance 
of  the  circuit.  Hence,  neglecting  capacity  and  leakage  conduct- 
ance, a  circuit  of  distributed  resistance  and  inductance  may  be 
considered  as  if  the  resistance  and  inductance  were  concentrated 
as  far  as  the  determination  of  the  current  is  concerned. 

Case  (a).  —  Unidirectional  voltage  impressed  on  the  circuit. 

For  the  current,  solution  of  (3)  gives 

E         f  Rl  +  Ri\ 

(1  —  €     LI+LI*     (see  equation  (17),  Chap.  I)    (5) 
i  \  / 


P7  ,    P 
til  -f  rt 

Substituting  (5)  in  (2) 


n 

\ 


From  (5)  and  (6)  it  is  seen  that  at  the  moment  of  closing  the 
circuit,  that  is,  when  t  =  0,  i  =  0  and  e  =  TJIT  1E.     In  the 

case  of  non-inductive  load  (when  LI  =  0),  e  =  jE  and  e  =  0  for 

x  =  0. 

It  is  interesting  to  note,  that,  while  resistances  consume  no 
voltages  when  i  =  0,  inductances  do  consume   voltages    when 

i  =  0,  provided  -^  =  0.     When  t  =  °°  ,  that  is,  when  the  current 
and  the  voltage  reach  their  permanent  conditions, 

77?  J?T      \       7? 

*'  =      ~        e  =      "    E'  the  expected  results- 


108  ELECTRICAL  ENGINEERING 

Problem. — Assume  reasonable  values  for  the  constants,  and 
plot  a  series  of  curves  for  the  voltages  at  various  values  of  t  and  x. 

Case  (b). — Alternating  voltage  impressed  on  the  circuit. 

From  equation  (39)  in  Chap.  I,  the  solution  of  (4)  is  found  to 
be: 

E  r  Rl  +  Rl  <t    ,  n 

i  =  |  [sin  (co*  -  13)  -  sin  (w*i  -  /3)e  "  £T+T7  (t  (7) 

where  ti  is  the  time  at  which  the  circuit  is  closed. 
Z  =  V(M  +  Ri)2  +  (LI  +  LJW 
and  _          ,  (LI  +  Li)  co 

(Rl  +  Ri) 
Differentiating  (7), 


cos  («/  -  0)  +  fj^-f^  sin  co/!  -  |9)e "  ^TiT  w       ' J  (8) 

Substituting  (7)  and  (8)  in  (2),  or, 

sin  (ft'  —  0)  sin  (wti  —  0) 

(9) 


where 

-Li)2 co2,  and  j8r  =  tan-1 


Referring  to  (7)  and  (9)  it  is  seen  that  the  current  is  the  same 
as  if  the  resistance  and  inductance  were  concentrated,  but  the 
voltage  is  different  at  different  points,  being  modified  in  magni- 
tude, and  displaced  in  time  phase. 

It  is  noticed  that  no  transient  component  in  the  voltage  or 
current  exists  at  any  point  of  the  line,  if  the  circuit  is  closed  at 

R 

ti  =  -,  or  in  other  words  when  sin  (wti  —  (3)  =  0. 

CO 

If  sin  (co/i  —  |8)  is  not  zero,  the  transient  voltage  appears  at  all 
values  of  x  except  x  =  I,  for  ft'  can  not  equal  (3,  or  — ^ — ZTT? — 

(LI  +  I/O  co 
can  not  equal  to     P7   , — 5 —  unless  x  =  I. 

£11  +  K\ 

When  t  becomes  large,  that  is,  many  cycles  after  the  circuit  is 
closed,  the  exponential  term  approaches  zero  and  the  whole  cir- 
cuit becomes  free  of  the  transient,  and  (7)  and  (9)  become : 

i  =  |  sin  (ut  -  |8)  (10) 

EZ' 

e  =  --sm  (co*  +  j8'  -  0)  (11) 


RESISTANCE  AND  INDUCTANCE  109 

Zf 

where  x  =  I,  that  is,  at  the  generating  end.  -^  =  I 

L 

&'  =  ]8,  therefore, 
e  =  E  sin  ut,  as  assumed. 

The  voltages  at  other  values  of  x  can  readily  be  computed 
from  (11)  and  it  is  seen  at  once  that  the  amplitude  is  proportional 

7' 

to  -;=  and  the  phase  is  ($'  —  /3)  radians  leading  the  impressed  e.m.f. 

The  effective  value  of  the  voltage  at  any  value  of  #  is: 

(12) 

where  ee/f  =  the  effective  value  of  the  voltage  at  x,  and  Eef/  that 
at  generator,  that  is,  at  x  =  I. 

When  the  line  is  open,  that  is,  when  Ri  =  oo ,  then  (7)  and  (9) 
become :  i  =  0,  and  e  =  E  sin  co£  for  all  values  of  x.  No  current 
flows,  which  is  to  be  expected. 

Grounding  the  receiving  end  of  the  line,  Ri  =  0  and  LI  =  0. 

.'.  -jj  =  -y  and  &'  =  j8,  hence  (7)  and  (9)  become: 
/         i 


i  =   2£77[sin  (ut  -  0")  -  sin  (wti  -  /3")e     £v      "'J       (13) 

and  Ex   . 

e  =  -£•  sin  at  (14) 

where  „.,          / ,  ,.,,  _,  Lo> 

A     =  \/  R2  -f-  L2co  ,  and  p     =  tan     "5" 

It  is  interesting  to  note  that  in  this  case  the  voltage  has  no 
transient  component  and  is  in  time-phase  throughout  the  line. 

PROBLEMS 

1.  Assume  reasonable  constants  of  the  circuit  for  equation  (9)  and  plot  e 
against  t  for  (a)  x  =  0,  (6)  x  =  ^' 

2.  When  an  accidental  ground  occurs  on  an  aerial  transmission  line  the 
voltage  10  miles  away  from  the  generator  station  is  found  to  be  60  per 
cent,  of  the  generator  voltage.     Determine  the  point  of  grounding. 


CHAPTER  VI 

CIRCUIT  CONTAINING  DISTRIBUTED    LEAKAGE    CON- 
DUCTANCE AND  CAPACITY 

A  low-voltage  cable  may  be  considered  as  an  approximate 
representation  of  such  a  circuit,  since  it  contains  distributed 
leakage  conductance  and  capacity  but  usually  low  resistance  and 
inductance.  Since  the  resistance  and  inductance  are  considered 
negligible  as  a  limiting  case,  it  remains  to  consider  a  system  of 
parallel  conductance  and  capacity.  The  voltage  may  be  con- 
sidered the  same  at  all  points  of  the  circuit,  that  is,  independent 
of  x. 

Let  i  in  Fig.  51  be  the  current  at  x,  then  i  +  ---dx  is  the  current 

at  x  +  dx.     Let  C  be  the  capacity  in  farads  per  unit  length  of  the 
conductor  against  the  ground  or  neutral,  and  G  the  conductance 


'i 

-hdz 

-> 

- 

-,* 

1 

1 

> 

1 

J 

1 

r\ 

=    : 

i  - 

--   : 

_       < 

>  ~ 

< 
< 
* 

•>  - 
> 

1- 

> 

< 

!    i 

>     ^ 

« 

:= 

>  - 

rr 

\ 

' 

1 

' 

FIG.  51. 

in  ohms  per  unit  length  of  the  conductor  to  the  neutral.     Hence 

de 
the  current  in  the  path  of  the  capacity  is  Cdx-?-,  and  the  current 

in  the  path  of  the  conductance  is  Gdxe.     The  difference  in  cur- 
rent between  the  two  sides  of  the  element  dx  is  —  dx.     Therefore 

dx 

T-  dx  =  Cdx  -JT  +  Gdxef 


or 


(1) 


110 


DISTRIBUTED  LEAKAGE  111 

This  equation  is  similar  to  (1)  in  Chap.  V  with  i  for  e,  e  for  i, 
C  for  L,  and  G  for  R. 

As  e  is  independent  of  x,  equation  (1)  integrated  gives: 

K  (2) 

It  is  of  no  interest  to  consider  short  circuit  of  the  cable,  since 
the  resistance  and  inductance  are  neglected,  for  it  would  mean 
a  dead  short-circuit  on  the  generator.  Therefore  consider  the 
case  of  switching  the  generator  on  the  open  cable.  Thus,  where 
x  =  0,  i  =  0.  .'.  K  =  0,  and  (2)  becomes: 

(3) 

Case  (a). — Unidirectional  voltage  impressed  on  the  cable. 
In  this  case,  it  is  assumed  that  e  =  E  from  i  =  0  to  t  =  °° , 
but  just  before  t  =  0,   e  =  0.     Therefore  it  is   assumed   that 

—  =  co  just  before  t  =  0,  and  -77  =  0,  just  after  t  =  0;  that  is, 

equivalent  to  assuming  that  the  fictitious  condensers  were  charged 
with  an  infinitely  large  current  during  an  infinitely  small  period. 
In  reality,  the  rise  of  the  impressed  e.m.f.  takes  time,  though 
extremely  short,  and  the  resistance  and  inductance  of  the  circuit 
limit  the  initial  value  of  the  current  and  lengthen  the  period  of 
charging. 

These  assumptions  thus  do  not  allow  a  study  of  the  transient 
condition.  The  equation  indicates  simply  that  i  •=  <*> ,  for  t  =  0. 

de 
For  the  permanent  condition  we  have  -rr  =  0,  and  (3)  becomes: 

i  =  GEx  (4) 

which  is  expected. 

Case  (b). — Alternating  voltage  impressed  on  the  cable. 

Let  the  impressed  e.m.f.  be  e  =  E  sin  wt,  and  the  time  of  apply- 
ing to  the  cable  be  fa,  thus  e  =  E  sin  u>(t  —  ti).  Hence, 

de 

-  -J7    =    Eu  COS  0>(2    —    ti). 

Substituting  these  in  (3) 

i  =  [Cw  cos  u(t  -  ti)  +  G  sin  w(t  -  ti)]Ex, 


112  ELECTRICAL  ENGINEERING 


or  i  =  #zVC2u>2  +  G2  sin  (ut  -  w«i  +  0)  (5) 

where  ft  =  tan-  g^ 

This  equation  represents  the  permanent  values  of  the  current, 
and  shows  that  the  current  is  proportional  to  x  at  all  values  of  t, 
and  leads  the  impressed  e.m.f.  by  /3  radians  at  all  values  of  x. 
The  transient  component  does  not  appear  in  this  equation,  as 
explained  in  case  (a). 

The  transients  will  be  studied  in  the  following  chapters,  where 
the  capacity  and  inductance  are  considered. 


CHAPTER  VII 

CIRCUIT  CONTAINING  DISTRIBUTED  RESISTANCE  AND 

CAPACITY 

In  the  study  of  the  problems  involving  distributed  inductance 
and  capacity,  and  the  simpler  problems  involving  the  penetra- 
tion of  current  or  flux  in  conductors,  etc.,  where  alternating  cur- 
rent of  sine  shape  is  assumed,  a  certain  differential  equation, 
given  below  is  met. 

Its  solution  is  of  importance  to  the  engineer  and  deserves 
consideration. 

The  equation  is 

^  -  *«  ^  (i) 

dx*~  '  at 

A  general  solution  which  can  readily  be  verified  differentia- 
tion is: 

y  =  AQ  +  2Aeax+bt  sin  (ax  +  #  +  7)  (2) 

or 
y  =  A0  +  Aie0lX+6"  sin  (aix  +  ft  it  +  71) 

+  A2ea*x+M  sin  (a2x  +  fat  +  72)  +  .     .     . 

The  evaluation  of  the  different  constants  is  accomplished 
partly  from  the  known  conditions  at  some  points  of  the  system, 
and  partly  by  solving  for  the  constants  by  differentiation  and 
substitution. 

In  most  problems,  y  or  its  derivative  is  known  at  some  point, 
where  for  instance  after  permanent  condition  has  been  reached 

y  =  Y  sin  ut. 

If  the  point  happens  to  be  where  x  =  x\,  then  equation  (2) 
becomes 

Y  sin  ut  =  A0  +  SAeoxi+w  sin  (aXi  +  fit  +.  T) 

=  Ao  +  SA'cM  sin  (#+/) 

=  Ao  +  A\eblt  sin  fat  +  T'i)  +  A'*** sin  fat  +  y'*)  +.    .    . 

113 


114  ELECTRICAL  ENGINEERING 

Writing  the  equivalent  of  sines  in  terms  of  e 

Jut  _  f-jwt  j(0it  +  y'i)   _ 

' 


Y- 


J(P*t  +  y',)    _   €-j(ft*t  +  y'd 

»  -£-  -+.    . 

it  -  j(0,t  +  y't) 


+  .    .    . 


Thus,  since  the  left-hand  member  contains  the  imaginary  only, 
and  the  right-hand  member  a  constant  and  the  complex  imagin- 
ary and  the  two  sides  must  be  equal  for  all  values  of  t  it  is  evident 
that  AQ  and  the  b's  are  separately  equal  to  zero. 

.'.  y=   2Aeax  sin  (ax  +  ft  +  7)  (3) 

^  =   2Aeax  [a  sin  (ax  +  ft  +  7)  +  a  cos  (ax  +  ft  +  7)] 
oX 

*~  =   2Aeax(a2  -  a2)  sin  (ax  +  ft  =  7)  +  2aa  cos  (ax  +  ft  +  7) 
oX 

^f  =   2Aeaxp  COS  (ax  +  #  +  7) 
at 

Substituting  these  values  in  (1)  and  equating  the  coefficients 
for  similar  trigonometric  terms, 

a*  -  a2  =  0,  and  2aa  =  kzfi  (4) 

.*.  (a  -\-  a)(a  —  a)  =  0,  thus  a  +  a  =  0,  or  a  —  a  =  0,  or  both. 

For  a  +  a  =  0,  or  a  =  —  a,  the  second  equation  of  (4)  gives, 

—  2«2  =  fc2/3,  which  is  evidently  impossible.     Thus  there  remains 

only  a—  a  =  0ora  =  a. 


Then  2a2  =  k2P,  or,  a  =  a  =  ±  fc  -v/^    where  a  and   a  must 

have  like  signs. 

The  general  solution  then  becomes: 

y  =  A,eax  sin  (ax  +  ft  +  71)  +  A2e~ax  sin  (  -  ax  +  ft  +  72)  (5) 
and, 

~  =  Aieaxa[  sin  (ax  +  #  +  71)  +  cos  (ax  +  ft  +  71)] 

-  Aze~axa[ sin  (  -  ax  +  ft  +  72)  +  cos  (  —  ax  +  ft  +  72)] 
=  V2  a  [A  ic0*  sin  (ax  +  ft  +  71  +  | ) 

-  Aae'^sin  (-  ax  +  #  +  72  +     )      (6) 


RESISTANCE  AND  CAPACITY 


115 


Application  of  these  equations  will  be  found  in  the  case  of  a 
circuit  of  distributed  resistance  and  capacity  but  negligible 
inductance  and  leakage  conductance,  such  circuit  being  approxi- 
mately represented  by  the  cable  in  Fig.  52. 

Let  R  and  C  be  the  resistance  and  capacity  respectively  per 
unit  length  of  the  cable.  Let  the  distance  be  counted  from  the 
receiving  end  of  the  line. 

Let  the  voltage  at  B  be  e,  and  the  voltage  at  A  e  -f-       dx. 

ox 

.'.  the  voltage  consumed  in  the  line  element  is: 


.  de  j       de  j 

e  +  —  dx  —  e  =  —  dx. 
dx         dx 


di 


Let  the  current  at  B  be  i  and  the  current  at  A  be  i  +  —  dx. 

ox 


Receiving  End 
of  Line 


A 

FIG.  52. 


Thus  the  difference  in  current  on  each  side  of  the  line  element  is 


. 

i  -f  —  dx  —  i  =  —  dx. 
ox  ox 


de 


or, 


.'.  —  dx  =  iRdx, 
dx 

de 

—  =  ^R 

dx 


(7) 


The  difference  in  current  on  each  side  of  the  element  is  the 
charging  current  of  the  element. 


or 


.    di    ,         nj  de 

.  .  —  dx  =  Cdx— 

dx  dt 

di  =      de 
dx          dt 


(8) 


116  ELECTRICAL  ENGINEERING 

di       1    d2e 
From  equation  (7)  we  get:  -  —  =  >,   ~~2 

oX         it   oX 

l_  d2e  _        d^ 

'  '  R  dx2  ~       di' 


or, 


d2e  de 


Referring  now  to  the  general  equation,  it  is  seen  that  k2  =  CR, 

[RC0 

and  y  =  e,  and  a  =  a  =  ±  \l~cT- 


.'.e  =  A^s'm&x  +  ft  +  71)  +  A2e~axsin(-  ax  +  ft  +  72)  (10) 

and, 

1  de       A/2a  r  .  /  TT\ 

*  =  Rdx=  -ir[^a*^(™+  fl  +  Ti  +  y)- 

A2e-°*sin(-  ax  +  ft  +  72  +  ^)]      (11) 

Case  (a).  —  Alternating  current  supplied  to  a  circuit  of  distrib- 
uted resistance  and  capacity. 

Example  No.  1.  —  If  the  voltage  at  the  generator  end  of  the 
line  (x  =  e)  is  e  =  E  sin  cot,  and  if  the  cable  is  open  at  the 
receiving  end,  then  i  =  0  for  x  =  0  and  all  values  of  t,  and 
e  =  E  sin  ut  for  x  =  e.  From  (10) 

E  sin  ut  =  Aieal  sin  (al  +  ft  +  7i)  +  A2€~al  sin  (-  al  +  at  +72) 
=  [Aieal  cos  (al  +  71)  +  A<*ral  cos  (-  al  +  72)]  sin  ft  + 

[Aieal  sin  (aZ  +  71)  +  A2€~al  sin  (-  al  +  72)]  cos  # 
.'.  Aieal  cos  (aZ  +  71)  +  A2e~al  cos  (—  al  +  72)  =  1? 
and  |8  =  w, 

and         Aiea/  sin  (aZ  +  7i)  +  ^2€-°z  sin  (-  al  +  72)  =  0        (12) 
For  x  =  0,  i  —  0  for  all  values  of  t  from  (11), 

0  =  Ai  sin  (#  +  71  +  j)  -  A2  sin  (#  +  72  +  | 

Since  this  must  hold  for  all  values  of  i, 

Ai  =  Az  =  A  and  71  =  72  =  7 
Then  from  (12)  we  have: 

E 


tal  cos  (al  +  7)  +  e~al  cos  (  -  al  +  7) 
and,  eal  sin  (al  +  7)  -f  e~°'  sin  ( —  al  +  7)  =0 


or, 


RESISTANCE  AND  CAPACITY 

f  =  tan"1  I    •    al        _al  tan  al    ,  and 


117 


A  = 


e  =  A  [eax  sin  (ax 
and. 


=  ^-~  A  [ea*  sin  ( 


E 

(13) 

2al   +  €-2aZ   +   2  (COS2  ttZ    -   SHI2  (rf) 

-f-  7)  +  e-°*sm  (-  a*  +  a>*  +  7)!        (14) 


(15) 


I  ax  +  a>c  +  7  + 

—  e~ax  sin  f  —  ax  +  a>2  +  7  + 
Where  A  and  7  are  given  in  (13)  and 

o-    +^/^ 

If  the  voltage  at  the  receiving  end  of  the  line  were  known  rather 
than  the  voltage  at  the  generator  then : 

e  =  EQ  sin  ut  for  x  =  0, 

EQ  being  the  maximum  value  of  the  voltage  at  the  receiving  end 
of  the  line. 

Thus, 
EQ  sin  ut  =  AI  sin  (/ft  +  71)  +  A2  sin  (/ft  +  72) 

=  sin  /ft(Ai  cos  71  +  A  2  cos  72)  -(-  cos  /ft(Ax  sin  71  -f  A2  sin  72) 
.'.  AI  cos  71  +  A2  cos  72  =  EQ,  fi  =  u  and,  AI  sin  71 

+  A2  sin  72  =  0     (16) 

For  x  =  0,  i  =  0  for  all  values  of  t  (assuming  again  an  open  line). 

.'.  AI  sin  \8t  -\-  7i  +  T)  =  A2  sin  (/ft  +  72  +  j) 

AI  sin  /ft  cos  (71  +  ^rj  -f  AI  cos  fit  sin  (71  +  T)  = 
A 2  sin  fit  cos  (72  H-  jj  +  A2  cos  fit  sin  (72  +  jj 

In  order  that  this  shall  hold  for  all  values  of  t,  the  coefficients 
of  the  similar  trigonometric  terms  of  t  must  be  the  same. 

.*.  AX  cos  (71  +  T)  =  A 2  cos  (72  +  T-J  and, 


or. 


AI  sin  (71  +  jj  =  A2  sin  (72  +  - 

.'.  tan  (7!  +  j)  =  tan  (72  +  | j , 
7i  =  7?  =  7- 


(17) 


118  ELECTRICAL  ENGINEERING 

Then  from  (16) 

(Ai  +  A2)  cos  7  =  EQ  and  (Ai  -f  A2)  sin  7  =  0 

.'.7  =  0  and  AI  +  A2  =  #o. 
From  (17)  A          ,          .       •   A  -  -°- 

^-1    —  /12  —   A        .  .   A    -        ~ 

Therefore 

771 

e    =    __°[€a*  gm   (aa.   _f_    ut)    +   €-ax  gm   (_    a:C 

and, 


where  a  =  ~\~  \l~~n  —  &nd  .E'o  is  the  maximum  value  of  the  e.m.f. 

at  the  receiving  end. 

In  the  examples,  both  1  and  2,  the  current  leads  the  voltage 
by  45°  at  all  points  of  the  line. 

Let  CQ  be  the  voltage  at  the  receiving  end  and  e\  that  at  the 
generating  end. 

From  (14) 

2E  sin  (co£  +  7) 
eQ  =  2A  sm  (wt  +  7)  =      , 

V>'  +  *-**  +  2  (cos2  al  -  sin2  al) 
and 

ei  =  E  sin 
From  (18) 

eQ  =  EQ  sin 
and 

TJ 

ci  =  ^[cai  sin  (al  +  «0  +  e~a'  sin  (-  aZ  +  ««)] 

TTf 

=  -n[(eal  +  €~°0  cos  aZ  sin  co^  +  (cai  —  e~al)  sin  ai  cos 


=   2^o     e2^  +  €~2^  +  2  (cos2  aZ  -  sin2  al)  sin  (co*  -  7). 

Hence  both  equations  show  that,  (a)  the  voltage  at  the  re- 
ceiving end  leads  the  generator  voltage  by  an  angle  7,  and  (6) 
the  maximum  voltage  at  the  receiving  end  is: 

_  2  _ 
V*201  +  e-20'  +  2  (cos2aZ  -  sin2aO 

times  the  maximum  generator  voltage.  In  fact  examples  (1) 
and  (2)  refer  to  one  phenomenon,  but  one  terminal  condition 
already  known  and  one  terminal  condition  to  be  determined  are 
interchanged  in  the  statements  of  the  examples. 


RESISTANCE  AND  CAPACITY  119 

Example  No.  3.  —  The  same  phenomenon  may  be  studied  in 
still  a  different  way,  namely,  measuring  x  from  the  generator 
end,  that  is,  x  =  I  refers  to  the  receiving  end  of  the  line. 

When  the  generator  is  taken  as  the  point  from  which  the  dis- 
tance is  measured,  then,  as  the  voltage  and  current  decrease  as 
x  increases,  we  have: 

—  —  dx  =  iRdx, 

C/Jv 

and  di   ,          „      de 

—  -—  dx  =  Cdx  — 
dx  dx 

which  by  a  similar  transformation,  also  resolves  in  the  differ- 
ential equation  : 

*_!?.  _  rp  de- 
dx*  ~      Kdt 

.*.  e  =  A^ax  sin  (ax  +  &t  +  71)  +  Ax~ax  sin  (-  ax  +  pt  +  72) 
and, 


For  x  —  I.     i  =  0  for  all  values  of  t. 

.'.  A**1  sin  (al  +  fit  +  71  +  j)  =  Ax~al  sin  (-  al  +  pt  +.  72  +  | 

or, 

=  A!€ai[sin  pt  cos  (al  +  71  +  ^)  +  cos  #  sin  (+  al  + 

=  A2e-ai|  sin  ^  cos  ^—  a^  +  72  +  ^j  +  cos  pt  sin  f  —  a/  +  72  +  4)  I 
As  this  must  hold  for  all  values  of  t, 

.'.  Aieal  cos  (al  +  71  +  ^)  =  A*-"1  cos  (-  al  +  72 
and, 

Ai6fll  sin  (a/  +  Ti  +  |)  =  A2c-ai  sin  (-  al  +  72  + 

.'.  tan  (al  +  71  +  ^j  =  tan  (—  al  -f-  72  -f-  |j 

/.  72  =  7i  +  2aZ, 
and  it  follows  that, 

At  =  Aie™  (20) 


120  ELECTRICAL  ENGINEERING 

For  x  =  0,  e  =  E  sin  ut  ; 

.*.  E  sin  cot  =  Ai  sin  (fit  +  71)  +  A2  sin  (j3t  +  72) 
=  (A  i  cos  71  +  A  2  cos  72)  sin  pt  +  (Ai  sin  71  +  A2  sin  72)  cos  fit. 

In  order  to  make  this  hold  for  all  values  of  t, 

|8    =    CO 

AI  cos  71  -f-  A  2  cos  72  =  E, 
and 

Ai  sin  71  +  A2  sin  72  =  0. 
From  (20) 

Ai  sin  7!  +  Ai€2oZ  sin  (71  +  2al)  =  0 

.*.  sin  71  =  —  €2aZ(sin  71  cos  2al  +  cos  71  sin  2aZ), 

-  e20*  sin  2aZ 
'  •  ^  =  tan      1  +  e-cos^ 

Let  7  =  71  +  al,  then  71  =  7  —  al,  and  72  =  7  +  aZ.     And 

let  A  =  Aie°z  =  A2e-°z.     Then, 

•pi 
e~al  cos  (7  -  al)  +  caZ  cos  (7  -f  aZ)  =  -T  (22) 

From  (21)  and  71  =  7  —  a£,  we  have: 

sin  (7  —  al)         —  e2aZ  sin  2al 
tan  71  =   tan  (T  -  =  COS(Y  _  oi)  -  j  +  €«  cos  2al> 


.'.  7  =  tan 


— i 


r-?l-(-«l 

_   e0(  +  e-ol 


which  is  the  same  as  that  in  example  (1) 
Substituting  the  value  of  7  in  (22) 


V  ^al  +  e~2ai  +  2(cos2  al  —  sin2  aZ) 

also  the  same  as  that  in  example  (1). 
Hence, 

e  =  A[e~a(l~x}sm(—  al  —  x  +  wt  +  y)  +  c^^^sin  (aZ  —  x  +  ut  -f  7)] 


sn    a 


which  are  identical  with  the  equations  obtained  in  example  (1), 
only  with  (Z  —  x)  in  the  place  of  x. 


RESISTANCE  AND  CAPACITY  121 

It  is  noticed  that  at  any  particular  point  of  the  line  the  current 
and  e.m.f.  waves  are  sine  waves. 


The  wave  length  X  is  found  when,  x\^~  = 

.*.  X  =   X  — i     =  =  2ir\hnr 

CRa)  \CRu 


mr/~      \fCR' 

The  time  required  for  the  wave  to  go  one  complete  wave  length 

isH- 

Thus  the  velocity  of  propagation  is: 

distance  /  w  I  Ifw  /2o> 

Thus  the  velocity  of  propagation  is  proportional  to  the  square 
root  of  the  frequency. 

Higher  harmonics  travel  faster  than  the  fundamental.  The 
third  harmonic  travels  73  per  cent,  faster,  etc. 

But  while  the  higher  harmonics  travel  faster  than  the  funda- 
mental their  attentuation  is  greater  as  will  be  seen. 

When  the  wave  has  traveled  one  complete  wave  length,  that 


is,  when  x  =  X  =  ^\J~( 

The  exponential  term  becomes: 


2 

CRu 


=  e~2v  =  4-  =  0.0019. 

That  is,  the  wave  is  only  0.2  per  cent,  of  its  original  value.     It 
has  reached  —  =  =  0.368  of  its  original  value  when 


CRu 

~~    =  l   °r  x  ~-= 


2  /     2  /I 

cT^  =  VCRW  =  \Cte 


Thus  the  third  harmonic  has  decreased  to  37  per  cent,  of  its 
original  value  in  a  distance  which  is  only  58  per  cent,  of  that 
required  by  the  fundamental  to  be  reduced  to  37  per  cent,  of  its 
original  value. 


122  ELECTRICAL  ENGINEERING 

To  find  the  time  for  the  wave  to  decay  to  -  of  its  original 
value,  we  have: 


distance          /     1  //TT 

time  =  — ; — r?—  =  -\/ ~^   „  -j-  2  A/T^:  = 
velocity        \CRirf  \CR 


Thus  the  time  required  for  a  given  decay  varies  inversely  as 
the  frequency.  The  third  harmonic  requires  only  one-third  of 
the  time  of  that  of  the  fundamental. 

Instance.  —  A  concentric  cable  100  miles  long.  Assume  a 
capacity  of  1  m-f.  per  mile  to  the  neutral. 

Using  the  mile  as  the  unit  of  distance, 

C  =  106. 

Assume  the  cable  to  have  a  resistance  (of  one  conductor)  of  1 
ohm  per  mile. 
Then  R  =  I 
At  60  cycles,  /  =  60  and  co  =  377. 

/2  X  377 
.'.  Velocity  of  propagation  =  A/     6  =  27.500  miles   per 

sec.     The  velocity  of  the  triple  frequency  wave  would  be, 

\/3  X  27,500  =  47.500  miles  per  sec. 

The  main  wave  is  reduced  to  37  per  cent,  of  its  original  value 
after  =  =  0.00265  sec.;  and  the  triple  frequency  wave  is 


reduced  to  the  same  fraction  in  one-third  of  the  time  or  0.0009 
sec.  In  the  first  case  the  wave  has  traveled  73  miles;  in  the  case 
of  the  triple  harmonic  42  miles. 

Problem.  —  Develop  the  equation  of  the  voltage  and  the  cur- 
rent in  a  closed  cable  under  alternating  impressed  e.m.f. 

Case  (6).  —  Direct  current  supplied  to  the  cable. 

Example  No.   1.  —  Consider   the    line    open  at  the  receiving 
end  (x  =  0). 
Assume, 

e  =  K  +  2Aax+bt  sin  (ax  +  ft  +  y) 
where  x  =  1,  e  =  E  for  all  values  of  t, 

This  is  evidently  only  possible  if 

2Aeal+bt  sin  (al  +  pt  +  7)  =  0  and  K  =  E. 


RESISTANCE  AND  CAPACITY  123 

As  this  must  hold  for  all  values  of  t,  p  =  0,  and  al  +  7  =  nir.1 

7T 

It  is  found  convenient  to  let  7  =   —  ~,  that  is,  to  make 

z 

cos  al  =  0,  and  al  =  mr  +      =   i^L+ii*: 


where  7        TT     STT     STT     7?r 

ai==  2  '  "2"    T    T'  etc" 


Thus,  e  =  E  +  ~SAtax  +  l"  cos 

f)    f* 


.   2  =  a    —  aeax        cos  ^  _  aaeax       sn 


ot 

Substituting  these  values  in  the  general  equation 
d*e  de 

.W     ™  Tt 

and  equating  the  coefficient  of  similar  trigonometric  terms  we 
get: 

a2  -a2 

~CR~ 

and  ax  =  0  or  a  =  0,  since  a.  can  not  equal  zero. 

'eosia+M?  (23) 


1  sin     v     '       '          (24) 
K  dX       K  n  =  o  2i 

When  *  =  0,  x  <  1,  e  =  0 


n  =  0 

1  In  this  equation  appear  several  constants,  some  of  which  are  determined 
by  the  terminal  conditions,  others  by  mathematical  transformations.  It  is, 
of  course,  possible  to  do  a  certain  amount  of  choosing  as  long  as  the  choice 
satisfies  the  differential  equation  as  well  as  the  known  conditions  which  exist 
in  the  problem.  So,  for  instance,  we  may  assign  an  arbitrary  value  of 
7  and  carry  the  calculations  through  when  we  may  find  that  the  final  expres- 
sion is  simple  or  too  complicated  to  be  of  practical  value. 

It  is  reasonable  that  in  the  first  trial  7  may  be  assumed  as  zero.  When 
the  problem  is  worked  out  on  this  basis  it  is  seen  that  the  answer  is  not  sus- 
ceptible to  a  simple  equation.  The  trial  will  suggest  another  value,  most 

likely  7  =  -  |.     This  is  therefore  used. 


124  ELECTRICAL  ENGINEERING 

In  order  to  determine  the  values  of  An,  multiply  both  sides 
of  (24)  by  cos  —   —  —    -  dx  and  integrate  between  0  and  e,  thus 


C 

Jo 


e       7r(l  +  2k}xn=~  .  7r(l  +  2n)z, 

cos—  ^  —  J—  —  -  2,    Ancos-    -7^7—   -  dx 
21          n=o  21 


Each  term  on  the  left-hand  side  equals  zero  except  that  one 
which  has  n  =  k,  and  hence  this  particular  value  is  used,  and  we 
have 


A,       cos*  *  <fc  =  -  E 


Integrated, 


coB  (25) 

TT   n  =  i  n  2/ 


• 
sin  -    -^--  (26) 

Lli  n  =  i  ^t 

The  voltage  at  the  receiving  end  is: 


For  i  =  0,  <  =  0  _  4£"""  (- 


,  TT2? 

which  is  zero,  in  accordance  with  the  assumption  made  in  develop- 
ing the  equation. 

Therefore,  incidentally,  we  get 


which  is  a  known  interesting  series  from  which  the  value  of  TT 
can  be  computed. 

The  current  at  the  generating  end  is 


(27) 
when  t  =  0  and  x  =  I, 


RESISTANCE  AND  CAPACITY  125 

which  is  a  limiting  value  never  reached,  since  with  the  slightest 
increase  in  t  the  series  converges  very  rapidly. 

7T2 

For  the  sake  of  briefness,  write  ra  for  ~p7^r2  then,  for  x  —  I, 

2E 

i  =  Jft(e~mt  +  e~9w'  +  *~25m'  +  e~49w<  +  .    .    .)  (28) 

From  (28)  it  is  readily  seen,  that,  when  t  has  any  appreciable 
value,  the  current  dies  out  approximately  according  to  the 
exponential,  €"«*'.  When  the  line  is  very  long,  the  initial  large 
current  will  remain  during  a  considerable  length  of  time.  When 
I  is  very  small,  the  limiting  case  is  that  of  concentrated  capacity. 
As  I  =  0  (28)  approaches: 

•-*-* 

HI  is  the  resistance  and  Cl  the  capacity  of  the  entire  line. 

In  the  case  of  concentrated  resistance  and  capacity  it  has  been 
shown  that 

E  -  l  t 


Comparing  the  equations  it  is  seen  that  the  transient  current 
can  be  fairly  well  approximated  by  assuming  that  the  line  capac- 
ity is  concentrated  in  the  middle  of  the  line. 

Example  No.  2. — In  case  the  line  is  grounded  at  the  receiving 

T^ 

end,  the  permanent  voltage  is  -T-- 

Thus,  _  Ex 

I 

7  may  in  this  case  be  conveniently  taken  as  zero,  thus, 

Ex  -^t 

-  2Ae     CR   sin  ax 


for  x  —  I,  e  =  E  for  all  values  of  t. 
Thus  sin  al  =  0,  and  al  =  rnr 

/.e=^  +  T^-"^'Sin^ 

1  n  =  0 

For  t  =  0,  e  =  0  for  all  values  of  x  <  L 

Ex 


sm  T  T 


126  ELECTRICAL  ENGINEERING 

An  is  determined  as  before  by  multiplication  and  integration 
and  we  get  finally: 


.         sin  (29) 

(30) 


J-l/lr  il/l/      n  _,  J 

The  current  at  the  receiving  end  is: 

¥1  =s  CD 


*     -, 

2  (31) 


CHAPTER  VIII 
DISTRIBUTED  INDUCTANCE  AND  CAPACITY 

Permanent  Condition. — Let  L  and  C  (Fig.  53)  be  respectively 
the  inductance  and  capacity  per  unit  length  of  the  circuit. 
The  voltage  consumed  in  the  line  element  dx  is: 

de    ,          ,  di  , 

— -  dx  =  L  —  dx  (1) 

ox  dt 


_J Receiving  End 

of  Line 


The  difference  in  current  between  two  sides  of  the  element  is : 


(2) 


Differentiating  (1)  with  respect  to  x 
d2e  di 

_    _,      j     ^ 

dx*  ~       dtdx 
Differentiating  (2)  with  respect  to  t, 

~dxdi  ==  C~di* 
d%_  _          M 

•'•  dx*         di 

dx~*  =  LCW 
127 


Similarly, 


(3) 
(4) 


128 


ELECTRICAL  ENGINEERING 


In  this  problem  there  is,  therefore,  encountered  an  equation  of 
the  following  type: 


= 
dt2  dx2 


(5) 


An  often  successful  procedure  for  finding  particular  solutions 
of  simple  partial  equations  of  this  or  similar  simple  types,  is  to 
assume  the  solution  to  be: 

y=UV  (6) 

Where  U  is  a  function  of  t  only  and  V  is  a  function  of  x  only. 
Differentiating,  we  get: 


— "  =  U 

dx2  dx2 

d2y  _  d2U 

dt2  ~  dt2 


Substituting  (7)  in  (5) 


^_^^Zor  J_^ 
dt2  "      U  dx2  °    Uk2   dt2 


1  d2V 

V  dx2 


(7) 


(8) 


Since  the  left-hand  member  is  a  function  of  t  only  and  the 
right-hand  member  a  function  of  x  only,  it  follows  that  each  side 
of  the  equation  must  be  equal  to  the  same  constant.  Let  that 
constant  be  —  a2. 


and 


d'V  2T7 

— -  =   —  a2V 
dx2 


(9) 


The  following  trigonometric  terms  evidently  satisfy  (9) . 

U  =  sin  akt  or  U  =  cos  akt 
V  =  sin  at    or  V  =  cos  at 

Thus  the  solution  is: 

y  =  K  -j-  S[Ai  sin  ax  sin  afcZ  -f-  A  2  sin  ax  cos  fccrf  + 
As  cos  ax  sin  aAtf  +  A  4  cos  ax  cos 

Where  AI,  A  2,  A3,  A4  and  K  are  to  be  determined,  and  k  = 
and  the  S  sign  refers  to  summation  with  all  possible  values 


(10) 


(U) 
1 


of 


a. 


INDUCTANCE  AND  CAPACITY  129 

Consider  now  the  specific  case  of  an  open  alternating-current 
line  of  negligible  resistance  and  leakage.  Determine  the  values 
of  the  current  and  e.m.f.  at  any  time  at  any  point  of  the  line  after 
the  permanent  condition  has  been  reached. 

If  the  distance  is  counted  from  the  generator  end  the  generator 
voltage  is  e  —  E  sin  ut 

then  di  de 

-dxdx  =  +Cdidx 

and,  de  di 


The  final  differential  equation  becomes,  the  same  as  equation 
(3). 
The  conditions  for  open  line  are: 

for  x  =  0,  e  =  E  sin  ut 

for  x  =  I,  i  =  0  for  all  values  of  t. 

Since  we  are  dealing  with  permanent  condition  the  current  and 
e.m.f.  vary  with  fundamental  frequency  and  the  solution  is 
therefore  : 

e  =  A  i  sin  ax  sin  kat  +  A2  sin  ax  cos  kat  + 

As  cos  ax  sin  kat  +  A4  cos  ax  cos  kat 

i  =  A  5  sin  ax  sin  kat  +  A6  sin  ax  cos  kat  + 

A-j  cos  ax  sin  kat  +  A8  cos  ax  cos  kat     (15) 

These  are  related  by  the  equation: 

de  _  di 

dx~    ~      d~t 

„  -  =  a[A  i  cos  ax  sin  kat  +  A  2  cos  ax  cos  kat  — 

(j  30 

AS  sin  ax  sin  kat  —  A  4  sin  ax  cos  kat]. 

L  —  =  LkalAz  sin  ax  cos  kat  —  A6  sin  ax  sin  kat  + 
of 

A?  cos  ax  cos  fca£  —  A8  cos  a£  sin  fcorf]. 

Equating  the  coefficients  for  similar  trigonometric  terms,  of  t, 
AI  cos  ax  —  As  sin  a  =  —  Lfc[—  A6  sin  ax  —  A8  cos  ax]' 


and, 


cos  ax  —  A.4  sin  ax  =  —  Lk[Ab  sin  ax  +  A7  cos  ax] 


(A) 


130  ELECTRICAL  ENGINEERING 

Since  these  must  hold  for  all  values  of  x,  we  can  substitute 

ax  =  0  and  ax  =  ~> 
£ 

:.  Al  =  LkAB  (16) 

A2  =  -  LkA7  (17) 

and,                                  -A9  =  LAA6  (18) 

-A4  =     -  LkAb  (19) 
For  x  =  I,  i  =  0  for  all  values  of  t, 

.'  .  0  =  A  5  sin  al  sin  kat  +  A  6  sin  al  cos  fcotf  + 

A  7  cos  «£  sin  fccx£  -f  A.  8  cos  «£  cos  fc«£     (B) 

.'.A  5  sin  a?  +  AT  cos  a/  =  0 
.'.A  e  sin  al  +  A8  cos  al  =  0 

.'.A7  =  -  A5  tan  «Z  (20) 

/.A8  =  -  A6tan  al  (21) 

.'.  e  =  LkAs  sin  ax  sin  fcotf  —  LfcAr  sin  ax  cos  fc«Z  - 

LkA  6  cos  ax  sin  kat  +  LfcA  5  cos  ax  cos  fcaL 
For  a;  =  0,  e  =  E  sin  ut 

.'.  #  sin  ut  =  —  LkA&  sin  kat  —  LkA$  cos  /ca^, 

W 

.'  .  E  =  —  LkA&,  or  A6  =  —  yr-'  A5  =  0,  and  co  =  ka. 

From  (19)  and  (19)  A3  =  E 
and  A4  =  0 

From  (20)  and  (21)  A7  =  0 

W 

and,  A8  =  yy  tan  al 

From  (16)  and  (17)  Al  =  £7  tan  aZ 
and  A2  =  0 

Therefore,  A^  =  E  tan  al 

A2  =  0 

A3  =  # 

A4  =  0 

A6  =  0 

A 


A7  =  0 

A  8  =  TT  ^an  a^ 
i>/C 

e  =  E  tan  a?  sin  ax  sin  fcatf  +  E  cos  ax  sin 


INDUCTANCE  AND  CAPACITY  131 

'       i  =  --  v-r  sin  ax  cos  kat  +  -f-p  tan  at  cos  axkat. 
LK  LK 

Substituting,        ,   _        1  _  «  _ 

"  VLC  * 

e  =  #[tan  ul-\/LC  sin  co\/LC  £  sin  co£  +  cos  u\/LC  x  sin  coj], 

•rr 

v-~ 
or, 
e  =  E  sin  w<[tan  ul^/LC  sin  u\/LC  x  +  cos  u\/LC  x], 

/£  _  _ 

i  =  E-l~   cos  co^ftan  coZ\/LC  cos  u\/LC  x  —  sin  w\/EC  a; 


•rr 

t  =  v-~[tan  wl-\/LC  cos  w\/LC  x  cos  co£  —  sin  u^/LC  a;  cos  coZJ, 


or,  cos  co  \-LC  (Z  —  a;) 

e  =  #  sin  coi  -  ,  (22) 

cos  co  v  LC  t 


C  ^sin 

cos  coi  -  —  (23) 


The  voltage  at  the  end  of  the  line  is  : 

E  sin  co^ 


Example.  —  If  the  receiver  voltage,  instead  of  the  generator 
e.m.f.  is  known  and  if  the  distance  had  been  counted  from  the 
receiving  end  of  the  line,  then 

i  =  0,  for  x  =  0 

and,  de  _  di 

d~x~     'L  dt 

Thus  the  signs  for  A  5,  A6,  A^  and  A*,  in  equation  (A)  would 
have  been  reversed. 

:.Ai  =  -  AsLk. 
A2  =  +  A7Lk. 


A  4  =  — 
Equation  (B)  would  have  been: 

0  =  A  7  sin  kat  +  A  8  cos  kat,  .'.  A  7  and  A  8  =  0 
.'.  e  =  LkAs  cos  ax  sin  kat  —  LkA&  —  cos  ax  cos  fc 
For  c  =  0,  e  —  EQ  sin  ut. 

.'.  EQ  =  sin  co£  =  LkA6  sin  kat  — 
.'.  EQ  =  LkA6)  and  A5  =  0  and  o 
.".  e  =  EQ  cos  ax  sin  fca£ 

Z  =          s*u  ax  cos  ^a^ 


132  ELECTRICAL  ENGINEERING 

or,  e  =  EQ  cos  to\/LC  X  sin  coi  (25) 

i  =  Eo+Lr  sin  u\/LC  X  cos  WT  (26) 

\L 

Therefore  the  generator  voltage  is: 

e  =  cos  u-\/LC  IEQ  sin  cot  =  E  sin  co£, 

cos  co  \/LC  I 
and  (25)  becomes: 

cos  co  \/LC  x 

e  =  E  sin  coi  -  , ? 

cos  co  \/LCl 

which  is  obviously  identical  with  (23)  as  obtained  before. 
It  is  seen  at  once  that  the  receiver  voltage  is 

1 

cos  co  \/LC  I 

times  that  of  the  generator  e.m.f.  As  the  cosine  is  always  less 
than  unity  except  I  =  0,  the  receiver  voltage  is  always  greater 
than  the  generator  e.m.f. 

Therefore  the  receiver  voltage  would  approach  infinity,  when 


2wfVLCl  =  I 


,       J_  _1 JL 


_  

4(Ll)(Cl)         4L0  Co 

that  is,  when  the  natural  frequency  of  the  line  and  the  frequency 
of  the  impressed  e.m.f.  coincide. 

The  wave  length  is  X  =  — / 

WV.L/O  Co 

X  2^/  1 

Thus  the  velocity  of  propagation  =  —  =        /,     „-  ==     /,    ^ 

1        coV^o^o        V-^o  GO 

If  the  inductance  inside  of  the  conductor  is  negligible,  then  the 
velocity  becomes  that  of  light  =  188,000  miles  per  second.  In 
reality  it  is  somewhat  less. 

So  for  instance  in  a  transmission  line  consisting  of  No.  0 
B.  &  S.  wires,  18  in.  apart, 

L  =  1.6  X  10~3  henrys  per  mile. 
C  =  0.019  X  10~6  farads  per  mile. 

-4=  =  182,000  miles  per  sec. 


INDUCTANCE  AND  CAPACITY  133 

For  short  distances, 

sin  co  -\/LC  x  =  u-\/LC  x 
cos  co  \/LC  x  =  I 
.*.  e  =  Eo  sin  ut 


Ic 

i  =  Eo  -^1     co  -\/LCx  cos 


jC/o 

cos  co£  =  —  cos  co£ 

Xc 

where  xc  is  the  capacity  reactance  of  length  a  of  the  cable.  It  is 
seen  that  the  current  in  time  phase  leads  the  voltage  by  90°. 

Transient  Condition. — When  a  steady  voltage  is  impressed 
upon  the  circuit. 

DR.  FRANKLIN  in  his  book  on  waves  and  his  paper  before  the 
A.  I.  E.  E.  of  April,  1914,  has  approached  the  subject  from  a  most 
simple  and  instructive  point  of  view  and  has  been  able  to  make 
some  generalizations  which  are  of  great  value. 

He  shows  that  whatever  the  distribution  of  the  current  or 
e.m.f.  in  a  travelling  wave  along  a  transmission  line  there  must 
be  a  fixed  ratio  between  the  instantaneous  values,  which  ratio  is 

C 

Y  when  the  line  resistance  and  leakage  reactance  are  negligible, 

and  it  can  be  represented  by  a  somewhat  more  complicated  ex- 
pression when  they  are  taken  into  consideration. 

His  reasoning  is  briefly  as  follows: 

If  the  current  in  an  element  of  the  line  is  i  the  magnetic  flux 
in  the  area  a,  b,  c,  d,  Fig.  53,  is  Lidx. 

If  the  current  wave  progresses  toward  the  right  with  a  velocity 

V  the  time  required  for  the  flux  to  sweep  past  be  is  -y]  thus  the 
e.m.f.  induced  along  be  is  ^x  =  LiV. 

y 

Similarly  if  e'  is  the  voltage  in  the  line  element  then  the  charge 
on  ab  is  e'Cdx. 

This  charge  flows  past  the  point  in  time  y,  where  V  is  the 

velocity  of  propagation  of  the  e.m.f.  distribution;  thus 
.,  _  Q  _     Cdx 
1         t=   ''   dx    :=CeV' 
V 


134  ELECTRICAL  ENGINEERING 

In  order  then  that  these  distributions  shall  sustain  each  other, 
i  =  {',  e  =  er  and  V  =  V. 

•* 

.'.  e  =  LiV  and  i  =  CeV 
or 


and 


Ce'  -  Li-  or,^  =  +         - 

e          '  \L 


The  -f  sign  belonging  to  outgoing  waves,  arid,  the  —  sign  to 
the  reflected  waves. 

.  i  1C       ,  i'  1C      i          i' 

.  .  -  =  +  A/T  and  -SB  —  - /--  or-=  —  - 

e  \L         e  \L      e  e 

where  index  '  refers  to  the  reflected  waves. 

When  the  line  is  open  at  the  receiving  end  the  sum  of  the  in- 
coming and  reflected  current  waves  must  be  zero,  thus  i  -\-  i'  =  0 

,  .,  e  i  .  c'  =  +  e 

.  .  e'  =   —  i'  -.  =  -f-  -e  =  +  e     . 
i  i 


When  the  line  is  short-circuited  at  the  receiving  end,  e  +  e'  =  0 
.'.  ef  =  —  e.     Thus  i'  =  i. 

When  the  receiving  circuit  is  noil -inductive  and  of  resistance/^, 

e  +  e'  =  R(i  +  i'} 
but  e          1L  e 


substituting  these  values  above,  then, 

(R  -  a)  .  a  -  R 

e    =  e  ~^r-t and  i    =  i  —   — ^* 

R  +  a  a  +  R 

It  is  seen  that  the  reflected  current  and  e.m.f.  waves  may  be 
positive,  zero,  or  negative,  depending  upon  the  relative  values 

of  R  and  .*/- 

In  DR.  FRANKLIN'S  American  Institute  Paper  (April,  1914)  is 
given  a  very  full  discussion  of  the  nature  of  these  reflected  waves 
and  some  highly  instructive  diagrams  are  shown. 

For  example,  when  the  receiving  circuit  is  inductive  the  line 
acts  at  the  instant  of  reflection  as  if  it  were  open  circuited,  since 
the  current  can  not  rise  instantaneously  in  an  inductive  circuit. 
After  some  time  the  condition  becomes  that  of  a  non-inductive 


INDUCTANCE  AND  CAPACITY  135 

receiving  circuit,  discussed  above  (since  we  are  dealing  with 
direct-current  voltage).  Between  the  two  periods  of  time  the 
current  and  e.m.f.  change  according  to  a  simple  exponential 
law. 

Of  special  interest  is  the  condition  of  the  waves  when  the  line 
constants  change.  DR.  FRANKLIN  illustrates  this  condition  in 
the  case  of  an  overhead  line  connected  to  a  cable  system. 

Let  iy  ir  and  it  be  the  instantaneous  values  of  the  outgoing 
current,  the  reflected  current  and  the  transmitted  current,  and 
let  e,  er  and  et  be  the  corresponding  values  of  the  e.m.f. 

Then  e  +  er  =  et 

i  +  ir  =  it 

e  er          et 

*  -  =  —  -  =  a  ~  =  b 

±+  I  Ir  It 

From  these  equations  are  found 

b  -a 


It  is  of  interest  to  apply  these  simple  relations  numerically. 
Assume  that  the  inductance  and  the  capacity  of  a  cable  sup- 
plying power  to  an  overhead  line  are:  L  =  0.0002  henrys  and 

0  8 
C  =  ~Q  farads  per  mile,  and  that  the  corresponding  constants 

for  the  overhead  line  are  LI  =  0.0015  henrys  and  C\  =  ~T^ 
farads  per  mile. 


dllCl  -,  ;  \J*\J\J1.U  /, ~      „„„  v_    i 


If  therefore  such  cable-overhead  line  combination  is  connected 
to  a  source  of  steady  e.m.f.,  e,  the  voltage  at  the  junction  as  the 

548 
wave    reaches   it    will    be    et  =  oo?  o  =  1-88  times  that  at  the 

.6OC7.O 


136  ELECTRICAL  ENGINEERING 

generator.  Should  the  overhead  line  be  open  at  the  receiving 
end  the  voltage  will  be  doubled  as  the  reflected  wave  starts  on 
its  journey  back.  Thus  as  a  maximum  at  the  junction  the  volt- 
age would  equal  3.76  times  the  impressed  value. 

The  mathematical  solution  of  the  problem  is  given  in  equation 
(11)  which  can  be  written  in  the  following  way: 

e  =  K  +  ZA  sin  (  ±  ax  +  kat  +  7)  (27) 

where 


= 


+  a  applies  to  the  waves  issuing  from  the  generator  and  —a 
to  those  going  toward  it.  From  the  expression  ±  ax  +  kat,  it 
is  seen  that  the  waves  of  all  frequencies  travel  with  the  same 
velocity,  +k  or  —  k  where  the  signs  indicate  the  direction  of 
motion. 

It  will  be  shown  that  in  the  case  of  an  open  line  connected  to  a 
source  (of  negligible  resistance)  of  undirectional  voltage,  four 
waves  have  to  be  considered  before  the  cycle  repeats  itself. 

First  the  outgoing  rectangular  wave  of  value  E  which  begin- 
ning at  the  generator  progresses  toward  the  open  end  of  the  line. 
Second  the  reflected  wave  also  of  strength  E  which  returns  from 
the  open  end  toward  the  generator  which  with  the  initial  wave 
gives  a  wave  of  double  voltage.  Third  a  negative  wave  of 
strength  —  E  which  progresses  from  the  generator  toward  the 
open  end  of  the  line,  which  wave  is  necessary  in  order  to  maintain 
the  generator  voltage  E.  Fourth  the  reflected  wave  of  the 
negative  wave  which  is  of  strength  —  E  and  which  progresses 
toward  the  generator. 

Consider  now  what  happens  at  a  point  located  say  at  one-fourth 
of  the  length  of  the  line  from  the  generator. 

If  the  time  required  for  the  wave  to  reach  the  end  of  the  line 
is  T,  it  is  evident  that  during  Y±  T  there  is  no  voltage  at  the  point. 
After  that  time  the  voltage  remains  constant  at  a  value  E  until 
the  first  reflected  wave  arrives.  This  occurs  evidently  when  t  = 

1%T.     Thus  between  t  =  ^  and  t  =  1.7577  the  voltage  at  the 

point  is  E. 

From  that  on  it  has  a  value  of  2E  until  the  negative  gener- 
ator wave  reaches  the  point  which  occurs  when  t  =  2T  -f- 

T 

-     =  2.25  2T.     After  that  time  the  voltage  has  a  value  of  2E  - 


INDUCTANCE  AND  CAPACITY 


137 


E  =  E  until  the  reflected  wave  of  the  negative  wave  arrives 

T 

which  is  when  t  =  4T  -  -j  =  3.75 T7.     Then  the  voltage  =  2E  - 

2E  =  0,  and  it  remains  zero  until  a  time  t  =  4.25T7  when  the 
voltage  again  equals  E  and  the  cycle  is  repeated. 
The  result  is  the  wave  shown  in  Fig.  54. 

E.M.F,  Wave 


E 

',., 

5          1 
2 

.0        1 

5 

2 

2(1 

0 

-X) 

2 

5        3 
2 

0        3 

X 

.5 

J.O 
2(1  -x) 

4 

5 

k 

A 

V 

fc 
Q. 

r,i 

A 

k 

A  train  of  waves  would  pass  the  point  indefinitely  since  we 
have  neglected  the  energy  loss  in  resistance.  The  wave  length 
is  evidently  four  times  that  of  the  open  line. 

Consider  now  the  current  wave  of  Fig.  55. 

As  successive  equal  elements  of  the  line  are  being  charged  to 
voltage  E  a  constant  current  has  been  shown  to  flow  from  the 
generator  while  the  voltage  wave  progresses  toward  the  end  of 

Current  Wave 


l-x 

Jc 

1+ 

X 

k 

3i- 

X 

** 

t 

Ti 

31  4 

X 

k 

-J 

FIG.  55. 

the  line.  At  the  end  the  current  must  be  zero,  therefore  the 
reflected  current  wave  must  be  equal  but  opposite  to  the  incoming 
wave.  The  reversed  current  reaches  the  generator  after  a  time 
2T,  when  the  current  becomes  zero.  After  that  time  the  genera- 
tor supplies  —  E  voltage  and  a  negative  wave  of  current  flows 
until  it  also  is  neutralized  by  the  reflected  current  which  occurs 
when  t  =  4T. 

Consider  the  current  at  the  particular  point  mentioned  above. 


138  ELECTRICAL  ENGINEERING 

T 
From  t  =  0  to  t  =  -j  no  current  flows.     After  that  the  current 

is  constant  until  the  reflected  current  reaches  the  point  (at  t  = 
1.75T)  when  it  drops  to  zero.     It  remains  zero  until  the  negative 

current  issuing  from  the  generator  reaches  the  point  fat  t  =  2T 

T\ 
-f  -T)  .     Then  it  becomes  negative  and  remains  negative  until  the 

negative  reflected  current — now  positive — reaches  the  point  It  = 

T\ 
4T  —  -j]t  when  it  again  is  zero — and  so  forth. 

In  general  centering  our  mind  on  a  particular  point  x,  from  the 
receiving  end  of  the  line  there  is  no  e.m.f.  or  current  at  that  point 

until  t  =  — JT — .     After  this  the  voltage  is  E,  the  generator  voltage 

for  some  time.     At  t  =  T,  the  waves  reach  the  end  of  the  line  and 

I  +  x 
reflect,  therefore  after  t  =  — j- — ,  e  =  2E  for  a  period  of  time.     At 

21 
t  =  -j-,  the  waves  return  to  the  generator  end.     In  order  to  keep 

the  voltage  at  the  generator  end  constant  at  E  the  generator  must 
now  begin  to  supply  —E.     Therefore  after  time  t  =  — ]T~~>  e  at  x 

becomes  2E  —  E  =  E]  and  after  t  =  -W — ,  e  at  x  becomes  2E 

41 
—  2E  =  0.     At  t  =  -r  the  generator  reverses  its  voltage  from  —  E 

to  +E  again,  and  the  voltage  at  x  repeats  its  cycle  again  and 
again. 

Referring  now  to  equation  (27) 

e  =  K  +  SA  sin  (ax  +  kat  +  7)  +  SA'  sin  (-  ax  +  kat  +  7')    (28) 
when  x  =  I,  e  =  E  for  all  values  of  t. 

:.  2A  =  SA'  and  sin  (al  +  7)  =  -  sin  (al  +  7') 

*v'    —   'Y  ~4~  TlTT 

=  sin  (mr  —  al  +  7')  thus  a  =  -        ^ 

where  n  =  is  an  odd  number  (29) 

and  K  =  E 

:.  e  =  E  +  SA[sin(aa?  +  A;a/  +  7)+sin(-  ax  +  kat  +  y')]  (30) 


INDUCTANCE  AND  CAPACITY  139 

At  the  receiving  end  of  the  line  (x  =  0). 

e  =  0  for  all  values  of  t  which  are  less  than  T. 

But  when  t  =  T  the  voltage  is  2E. 
/c 

Thus  t  =  £  is  a  transition  point,  a  point  of  discontinuity,  e  is 

either  0  or  2E.     Substituting  the  two  values  in  equation  (30) 
we  get: 

+  E  =  2,4  [sin  (al  +  7)  +  sin  (al  +  7')]  respectively    (31) 

At  the  open  end  of  the  line  where  there  is  complete  reflection 
the  incoming  and  outgoing  waves  are  identical 

.'.  7  =  7'. 
Thus  from  (29)  a  =  -~j  where  n  is  an  odd  number. 

.'.  (31)  becomes  +  E  =  22A  sin  (—•  +  7)  (32) 

In  the  development  of  the  trigonometric  series  it  is  found  that  : 

4    n=c°  sin  nO 
+  1  =  -  -      S  -  where  n  is  an  odd  number        (33) 


where  the  negative  sign  refers  to  values  of  6  between  ?r  and  2w  and 
the  positive  sign  to  values  between  0  and  TT  or  TT  <  6  <  2ir  for 
negative  sign.  0  <  8  <  TT  for  positive  sign. 

See  "BYERLY'S  FOURIER'S  Series  and  Spherical  Harmonics" 
(page  51).  Comparing  equations  (32)  and  (33)  it  is  evident  that: 

_  E  4  1 

An  ~  2    IT  n 

and  nw 

y  +  T  =  nfl. 

It  remains  to  determine  the  value  of  6.  The  two  series  have 
TT  and  2-K  or  0  in  common.  It  remains  to  choose  the  proper  value 
of  these. 

If  TT  were  chosen  the  ±  signs  in  (32)  and  (33)  would  be  reversed, 
if  however,  0  or  2ir  is  chosen,  the  signs  are  satisfied.  Thus 

mr  nir 

+  7  =  2n?r  or  0     .  .  7  ==  - 


140  ELECTRICAL  ENGINEERING 

Thus  equation  (30)  becomes: 

e  =  E  +  j-  -S^Fsin  ^  (x  +  kt  -  1)  +  sin  ~  (-  x  +  kt  -  I)]    (34) 

a   Tt       H  L  £1  &l  J 


and  since  de          di^ 

dx  ~      dt 


2  i[sin  ^  (a  +  fc  -  0  -  sin  ^  (-  s  +  &  -  Z)]    (35) 
The  curves  drawn  in  Figs.  54  and  55  may  be  verified  as  follows  : 


= 


for  t  < 


x  =  T 
Z  -x 


k 

x  -\-  kt  —  I  <  x  -\-  I  —  x  —  I  or  smaller  than  0 
.'.  6  in  the  trigonometric  series  lies  between  TT  and  2ir. 

Therefore  S  -  sin  (a;  +  kt  -  Z)  =  -  £. 

Tl  4 

Consider  with  the  second  term  in  (34),  —x  +  kt—  I  <  —x+l 

31 
—  x  —  I,  thus  smaller  than  —  2x  or  smaller  than  —2  +  -r  or  —1.51 

thus  6  is  again  negative  and  the  series  of  the  second  term  in  (35) 
- 

—  -7  — 

kt  —  l  =  X-\-l  — 


adds  up  to  - 

.'.  e  =  E  -\-  ^  -  (—  -7  —  jj  =0  which  agrees  with  the  curve. 
When  k 

—       =        --       —       =        = 

we  are  in  the  first  quadrant 


and  -  x  +  kt  -  I  =  -  x  +  I  -  I  =  -  x  =  - 
6  lies  in  the  fourth  quadrant 

.*.  2J-  sin  (  —  x  -\-  kt  —  1)  =  —  2" 

.'.  e  =1£  which  agrees  with  the  curve. 
07 
For  j  =  _       k  +  kt-l=*x  +  2l  —  l  =  x  +  l=  1.751. 

6  lies  in  the  second  quadrant-  thus: 


INDUCTANCE  AND  CAPACITY  141 

and  -  x  +  kt  -  I  =     -  x  +  I  =  0.25Z. 
6  lies  in  the  first  quadrant,  thus  : 

Z^sin  (-x  +  kt  -  Z)  -  +£ 

TV  71 

.'.  e  =  2#  which  agrees  with  the  curve. 

The    current    wave    may    similarly    be    checked.     When    for 
instance  t  =  T,  it  is  readily  seen  that  the  algebraic  sum  of  the 

trigonometric  terms  become  j 

E  4    1C  TT  1C 

•  •  i  =  ?r  ~\  IT  o  =  E  \  T  which  agrees  with  the  curve. 

A      7T    \-L/     A  \  Li 


It  is  thus  seen  that  when  considering  the  outgoing  waves  only  the 

i          JC 

relation  between  the  current  and  e.m.f.  waves  must  be  -  =  \  T> 

e        \L 

the  equation  also  shows  that  when  considering  the  reflected  waves 


The  effect  of  the  line  resistance  is  to  taper  the  waves  so  that 
instead  of  their  being  represented  as  a  ribbon  of  parallel  sides 
the  sides  slant  toward  each  other;  thus  the  reflected  e.m.f.  wave 
is  not  as  great  as  the  original  wave,  and  the  line  soon  reaches  a 
state  of  permanent  condition. 

In  reality  the  wave  front  is  not  vertical  but  slants  and  the 
corner  is  rounded  off,  due  to  the  skin  effect  of  the  conductors. 
The  higher  harmonics  of  the  current  meet  a  much  higher  resist- 
ance than  do  the  lower,  and  hence  the  resistance  is  not  a  constant 
quantity  but  different  resistances  should  be  assumed  in  connec- 
tion with  the  different  harmonics. 

The  mathematics  involved  becomes,  however,  altogether  too 
complicated  for  any  practical  application.  The  important 
point  is  that  if  the  values  of  the  waves  are  determined  in  a  cir- 
cuit having  no  resistance,  the  most  pronounced  variations  in 
current  and  e.m.f.  are  discovered. 

A  circuit  having  no  resistance  and  no  leakage  is  said  to  produce 
pure  waves  the  characteristics  of  which  are,  as  has  been  shown, 
such  that 


142  ELECTRICAL  ENGINEERING 

That  is  the  electric  energy  is  always  equal  to  the  magnetic 
energy. 

The  wave  may,  however,  be  pure  even  if  there  is  resistance  and 
leakage  but  in  that  case  the  energy  dissipated  in  heat  per  unit 
length  of  line  must  be  equal  to  the  energy  dissipated  by  leakage 
in  the  electric  field. 

e2 
.'.  i2R  =  77-   where   R i   is    the    leakage    resistance    per   unit 

length. 

•    /?/?         e2       L 
'  '  RHl  ==  i*       C 

A  line  in  which  this  relation  exists  is  called  a  distortionless 
line. 

For  a  full  discussion  of  such  circuit  the  reader  is  again  referred 
to  DR.  FRANKLIN'S  book  on  waves. 


CHAPTER  IX 

DISTRIBUTED    RESISTANCE,    INDUCTANCE,    LEAKAGE, 
CONDUCTANCE  AND  CAPACITY 


Let  R,  L,  G  and  C,  Fig.  56,  be  the  line  constants  per  unit 
length  of  the  line,  K  being  expressed  in  ohms,  L  in  henrys,  G  in 
ohms  and  C  in  farads. 

The  voltage  equation  is  evidently 


or, 


or, 


de  .               di 
—  dx  =  Rdx  i  -\-  Ldx  -- 

ox  ot 

de  .           di 

dx  dt 

—  dx  =  Gdxe  4-  Cdx  TT 

ox  dt 


(1) 


(2) 


R      L 


FIG.  56. 

Differentiating  (1)  partially  with  respect  to  x  and   (2)   with 
respect  to  t  and  combining  the  results  with  (1)  and  (2)  we  get: 


^  =  LC  d~  +  (RC  +  GL)  ~  +  RGe 

OX"  Ol"  Ol 


and, 


(3) 
(4) 


where  a'  and  &  may  be  positive  or  negative,  real  or  imaginary, 
simple  or  complex. 

143 


The  general  solution  of  these  equations  is 


144  ELECTRICAL  ENGINEERING 

Substituting  the  general  solution  in  (3)  or  (4)  and  equating 
the  coefficient,  we  get 

a'2  =  LCp'*(RC  +  GL)ff  +  RG  (5) 

Substituting  a  +  ja  for  of  and  b  +  JP  for  ff  in  (5)  and  separat- 
ing the  real  and  imaginary  terms,  we  have 

a2  -  a2  =  LC  (62  -  02)  +  (RC  +  GL)  b  +  RG  (6) 

and  2aa  =  2LC6/5  +  (RC  +  GL)/?  (7) 

A  slight  consideration  shows  that  the  exponential  solution 
given  above  can  be  written 

e  =  k  +  ZAt*  ax  ±  bt  sin  (pt  ±  ax  +  7)  (8) 

If  now  for  the  sake  of  simplicity  only  the  permanent  condition 
is  considered  we  get 

e  =  k  +  SAc±ax  sin  (pt  ±  a*  +  7)  (9) 

If  as  a  further  limitation  the  current  and  e.m.f.  are  assumed 
to  be  simple  sine  functions,  depending  in  time  upon  the  impressed 
frequency,  then  p  has  only  one  value  <o.  From  (6)  and  (7)  follows 
then  that  only  two  values  of  a.  and  a  exist,  one  being  positive  the 
other  negative 

.'.  e  =  Aitax  sin  (pt  +  ax  -f  71)  +  A2e~ax  sin  (pt  -  ax  +  73)   (10) 

In  this  equation,  one  term  represents  the  sum  of  the  outgoing, 
the  other  the  sum  of  the  incoming  waves. 

If  the  line  is  open  at  the  receiving  end  then  the  beginning  value 
of  the  reflected  waves  must  be  identical  with  the  final  value  of 
the  incoming  waves  when  x  =  0. 

Thus  under  this  condition  for  x  =  0 

Al  sin  (pt  +  71)  =  At  sin  (pt  +  72) 
Since  this  must  hold  for  all  values  of  t 

•  '•  7i  =  72  and  A\  =  A% 

If  the  voltage  at  the  generator  end  is  E  sin  o>/,  then 
E  =  sin  ut  =  A1  [eal  sin  (pt  +  al  +  71)  +  t~al  sin  (pt  -  al  +  7^] 
which  by  simple  transformation  becomes 

E  sin  at  =  A  i  (sin  pt  [eal  cos  (al  +  7i)  +  e~al  cos  (-  al  -  71)] 
+  cos  pt[eal  sin  (al  +  71)  +  e~al  sin  (-  al  +  71)]} 


+  c-20'  +  2(cos2«Z  -  sin'aZ)  sin  (pt  +  B) 


DISTRIBUTED  RESISTANCE  145 

where      ,        _  _  eal  sin  (al  +  71)  +  e~al  sin  (  —  al  +  71) 
€al  cos  (al  +  71)  +  e~al  cos  (  —  al  +  y^ 

Thus  13  =  co  and  0  =  0 

# 

=  V^M-  <r2<"  +  27cos2 ~al  -  sin*  of) 
Since  0  =  0 

eal  sin  (al  +  71)  +  <ral  sin  (-  aZ  +  Tl)  =  0 

which  gives  eal  —  e~al 

71  =  tan"1 aT^~^.  ^an  °^ 

Equation  (10)  is  now  completely  determined. 
-Q-  +  Gc  =  Ai{Cw[eoa!  cos  (w«  +  0-0:  +  71)  +  e~ax  cos  (co/  —  ax  -f- 
71)  +  (7  [ea*  sin  (co^  +  ax  +  Tl)  +  €~oa;  sin  (co£  -  ax  +  71] } 
=  AI  •\/C2oo2'  -|-  G2 [eax  sin  (co^  -f-  ax  -j-  71  ~h  ^)  ~h 

e-ax  gjn   ^^    _    ax   _|_   ^i    _|_    ^J     Qjj 

where  .  Cw 

<p  =  tan"1  ^  • 

Let  i  now  be  the  permanent  component  of  the  current,  and 
assume : 

di> 

—  =  Biaeax  sin  (cot  +  ax  4-  ^>i)  —  Bia*~ax  sin  (co£  —  «a 
ox 

+  Biaeax  cos  (co^  +  0:0;  +  <^?i)  —  Biat~ax  cos  (coi  —  ca 


z  sin  (co£  —  ax  +  <^2  +  <r)      (13) 

,  « 
"1  -• 
a 

According  to  (2),  (11)  and  (13)  must  be  identical,  and  hence, 


where  ,  « 

<r  =  tan"1  - 
a 


Pi  -f  <r  ==  71 

To  sum  up, 

E'  [  eai  sin  (co^  +  ax  +  <pt)  +  €~oa;  sin  (ut  -{-  ax  +  71)] 
A/c2ai  +  €~2al  +  2  (cos2  al  -  sin2aZ) 


(15) 


'-*>pyf? 

[  eaz  sin  (w^  +  aa;  +  71)  —  e~oa;  sin  (coi  —  ao:  + 
V^  +  e~2al  +  2  (cos2  al  -  sin2^) 


146  ELECTRICAL  ENGINEERING 

where 


[eal   —   f—al 
^r+7^  tan  a 


1 

a     =  tan"1 
a 

.  Ceo 
<p     =  tan-1 


(Pi  =  <P  +  7i  — 
and, 


a  = 


+ 


the  latter  two  values  being  determined  from  (6)  and  (7)  by  let- 
ting 6  =  0  and  (3  =  w, 

These  solutions  apply  when  the  transient  terms  become  negli- 
gible, i.e.,  when  t  is  large  enough  to  make  e~bt  comparatively 
small. 

Case  (6).  —  Direct-current  distribution  in  an  open  line.  Con- 
sider the  line  open  as  before.  For  the  permanent  component  of 
the  solution,  i.e.,  a  solution  which  applies  after  the  line  has  been 
switched  to  the  generator  for  a  sufficient  length  of  time,  the  equa- 
tion can  be  derived  as  follows: 

Referring  to  equations  (6)  and  (7),  b  is  zero,  when  only  the 
permanent  component  is  considered,  and  ft  is  also  zero,  as  there 
exists  no  periodic  phenomenon,  when  the  impressed  voltage  is 
constant  and  when  the  starting  phenomenon  is  reduced  to  negli- 
gible magnitude. 

Substituting  6  =  0  and  ft  =  0  in  (6)  and  (7)  we  get: 

a2  -  a2  =  RG  and  2acx  =  0, 

from  the  latter,  either  a  or  a  must  be  zero,  while  from  the  former 
a  can  not  be  zero,  since  a  itself  must  not  be  imaginary. 

/.  a  =  0  and  a  =  ±  \/RG. 
•Let  ef  be  the  permanent  component  of  the  voltage  and  assume: 

er  =  A^x  +  A2e~ax  (18) 

where  x  =  I,  e'  =  E 

l  (19) 


DISTRIBUTED  RESISTANCE  147 

Let  i  be  the  permanent  component  of  the  current.     According 

to  (2), 


Substituting  (18)  and  (20) 

£-•• 

Integrating,        ^  =  G  ^ox  _  A^_ax^  +  R  (21) 

According  to  (1)  de'  .          di' 

Tx~  Rl+L~dt 

Substituting  (18)  and  (21)  in  (2), 

—  (A  €«*  -  A  e— ) 
a 

Since  a2  =  RG,  K  =  0,  and  (21)  becomes: 

I---WI--A*— ) 

where  x  =  0,  ir  =  0,  .'.  AI  =  A2. 

From  (19)  .          .  E 

1  2    ~    €ai     I     e~ al 

Therefore,  ^  __  ^  e0'  +  €~aj; 

^  €j£-^  (24) 

where  a  =  +  *\/RG,  these  equations  apply  when  the  transient 
terms  become  negligible. 


10 


CHAPTER  X 

PERMANENT  CONDITIONS  WHEN  ONE  OF  THE  FOUR 
CONSTANTS,  R,  L,  G,  AND  C  IS  NEGLIGIBLE 

I.  R  =  0. 

Case  (a). — Alternating  current:  The  solutions  are  given  by  (14) 
and  (15)  in  the  previous  chapter,  but  in  this  case, 


a= 


andj 


Case  (6).  —  Direct  current:  Referring  to  (23)  and  (24)  in  the 
previous  chapter, 

a  =--  ^/RG  =  0 
/.  e'  =  E. 

Under  this  condition  the  equations  deduced  give  $  in  the  case  of 
the  permanent  current.  Thus  they  do  not  lend  themselves  to  the 
determination  of  the  current. 

II.  G  =  0 

Case  (a).  —  Alternating  current:  With 


a  =  +  \/-7r  1+  \/L26o2  +  Rz  —  Leo], 
and 


equations  (14)  and  (15)  in  the  previous  chapter  give  the  solution. 

Case  (6).— Direct  current:  From  (23)  and  (24)  in  the  previous 
chapter, 

e'  =  E  and  i  =  0. 

III.  L  =  0. 

148 


PERMANENT  CONDITIONS  149 

Case  (a). — Alternating  current:  In  this  case 


and 


1+  VG2  +  C2co2  -  <?]• 
Case  (6).  —  Direct  current: 


fax     I        -02; 

e'  -  E-- 

*  - 


and, 


€al  ~  6~ 

IV.  C  =  0 

Case  (a).  —  Alternating  current:  In  this  case, 


and 


Case  (6).  —  Direct  current:  Same  as  III. 


CHAPTER  XI 

THE  DISTRIBUTION  OF  FLUX  OR  CURRENT  IN  A  CYLIN- 
DRICAL OR  FLAT  CONDUCTOR 

The  general  reasoning  and  the  mathematics  involved  in  the 
study  of  flux  or  current  distribution  in  conductors  is  very  simi- 
lar to  that  involved  in  the  study  of  propagation  phenomena  in 
transmission  lines.  It  is  therefore  included  in  this  part  of  the 
book  even  though  it  is  again  and  more  fully  considered  in  a  later 
chapter,  where  the  subject  is  approached  from  a  different  point 
of  view. 

Distribution  of  Flux  in  Cylindrical  and  Flat  Bars. — When  a 
cylindrical  bar  is  magnetized  by  a  winding  surrounding  it,  the 


etizing  Winding1 


FIG.  57. 

flux  of  final  flux  density  corresponding  to  the  external  m.m.f. 
appears  at  the  surface  nearest  to  the  magnetizing  winding. 

At  a  distance  from  the  surface  of  the  bar,  the  flux  density  is 
less  than  that  at  the  surface,  because  as  the  flux  penetrates  the 
inner  layers  of  the  bar,  it  induces  a  voltage  in  the  outer  layers, 
which  causes  a  flow  of  current  that  produces  m.m.f.  of  a  direction 
more  or  less  in  opposition  to  the  external  impressed  m.m.f. 

Referring  to  Fig.  57,  consider  a  concentric  tubular  element  of 

150 


DISTRIBUTION  OF  FLUX  151 

thickness  dx  and  mean  radius  x,  then  another  of  thickness  dx  but 
mean  radius  x  +  dx. 

Let  <{>  be  the  flux  in  the  tubular  element  of  radius  x,  and  <f>  +  d<fr 
that  of  radius  x  +  dx.  Thus  d</>  is  the  increment  of  flux  in  the 
tubular  element,  as  x  increases  from  x  to  x  +  dx,  but  the  total 
flux  in  the  tubular  element  is  0. 

</>  is  the  result  of  the  external  m.m.f  .  and  the  m.m.f.  (demagnet- 
izing) due  to  the  current  between  x  and  XQ',  </>  +  dcf>  is  the  result 
of  the  external  m.m.f.  and  the  m.m.f.  due  to  the  current  between 
x  +  dx  and  XQ.  Therefore  d<j>  is  caused  by  the  decrement  of 
demagnetizing  m.m.f.  due  to  the  current  between  x  and  x  +  dx, 
i.e.,  within  dx. 

Let  i  be  the  current  density  at  x,  then  the  current  within  dx  is 
ildx,  and  the  m.m.f.  due  to  it  is  also  ildx,  as  the  number  of  turns 
is  unity  (7  being  the  length  of  the  cylinder). 

Let  B  be  the  flux  density  at  x  then  dB  the  increment  of  flux 
density  as  x  increases  from  x  to  x  -\-  dx.  Thus  d(j>  =  2irxdxdB. 

m.m.f. 


Since  flux  =  0.4  TT 


and  the  reluctance  in  this  case  is 


—  -=  — 
reluctance 

I 


2ir  X  dxu,  * 


OAirildx 

We  get  d<p  =  2>jr  X-d  X  dB  =  -    —,— 


thus  dll  =  .  ,^ 

If  p  is  the  spec,  resistance 

then  the  resistance  that  the  current  within  dx  meets  is    I     7.7 

and  the  e.m.f.  consumed  by  the  resistance  =  ildx   I  yr~  =  2-jrpxi. 

Let  e  be  the  e.m.f.  induced  in  the  circle  of  radius  x,  and  e  +  de 
that  in  the  circle  of  radius  x  +  dx. 

As  no  external  e.m.f.  is  applied  around  the  circle  of  radius  x 
the  sum  of  the  consumed  and  the  induced  e.m.fs.  is  zero,  thus: 

e  +  Zirpxi  =  0  (2) 

Substituting  (2)  in  (1) 


UlJ  v/.i/iyuc/  /Q\ 

£•  "  -  ^—r  W 


152 


ELECTRICAL  ENGINEERING 


e  is  induced  by  all  the  flux  within  the  circle  of  radius  x,  and  e  -f- 
de  by  all  the  flux  within  the  circle  of  radius  x  -f-  dx,  thus  de  is 
induced  by  all  flux  in  the  tubular  element  2irxdx,  which  is  0 

1     d<t> 
according  to  our  notation.     Hence  de  =  —  ^cs  -77  or  using  partial 

differentials, 


de  - 

108  dt 


or 


or, 


0.2-n-xdxdB 


de 


108          dt 

2irx  dB 

io8  l>i 


(4) 


Equation  (3)  may  be  written: 

—  = 

Differentiating  with  respect  to  x, 
d2B      dB 

X   — r  -f   — -   =    — 

dx2        dx 
Combining  (4)  and  (5),  dividing  by  x 

d2B       I  dB       0.47rju 
dx2"  +  x  dx  = :  T08P 


Q.47TM  de 


+x> 


dx 


dt 


(5) 


(6) 


A  long  thin  flat  bar  may  be  considered  as  a 
cylindrical  bar  of  infinitely  small  curvature  or 
infinitely  large  radius,  thus  x  considered  as  the 
radius  becomes  infinity  andeauation  (6)  becomes: 


d*B 
dx2 


0.47T/Z  dB 
108P     dt 


Wmdmg         while  dx  and  x  take  the  meanings  as  shown  in 
FIG-  58'        the  Fig.  58. 

Equation  (7)  may  be  directly  derived  from  consideration  of  a 
flat  bar  in  place  of  a  cylindrical  bar. 

Distribution  of  Current  in  Cylindrical  and  Flat  Bars. — Reason- 
ing as  in  the  previous  paragraph,  but  considering  the  current  and 
flux  interchanged  in  their  places,  not  only  similar  but  also  iden- 
tical equations  will  be  derived  for  the  distribution  of  current. 


DISTRIBUTION  OF  FLUX 


153 


Let  B  in  Fig.  59  be  the  flux  density  at  x,  i  be  the  current  density 
at  x,  and  i  +  di  the  current  density  at  x  +  dx. 
The  flux  in  the  tubular  element  dx  is  Bldx. 

The  reluctance  of  the  flux  path  is  —r-r- 

0.47T  m.m.f . 
Ihus  since  rlux  = 


reluctance 


2irBx 
m.m.f.  =  m  =  j^-r 

2  ATT p. 


(8) 


(Current 

I 

1 

Impressed 

i    1 

E.M.F. 

1 

FIG.  59. 

As  x  increases  from  x  to  x  +  dx,  the  m.m.f.  increases  from  what 
is  within  the  circle  of  radius  to  that  of  radius  x  +  dx 

dm  =  2irxdxi,  or  -T-  =  2irxi  where  i  is  the  current  density  at 


distance  x 

Differentiating  (8) 

dm 
dx 

Equating  (2)  and  (3) 


27T 


0.47TM 


(•£+') 


f. 

dx 


,  o.4,,,- 

x 


(9) 
(10) 

(11) 


As  x  increases  from  x  to  x  +  dx,  the  increment  of  current 
density  is  di,  and  the  increment  of  current  in  the  tubular  element 

is  2Trxdxdi.  ,  The  resistance  of  the  material  that  this  increment 

„/ 

Therefore  the  increment  of  the 


of  current  traverses  is 
consumed  e.m.f.  is: 


2irxdx 


2irxdxdi 


2-irxdx 


PUi. 


154  ELECTRICAL  ENGINEERING 

Hence  the  decrement  of  the  induced  e.m.f.,  —  de  is  —pldi.  This 
-de  is  caused  by  the  flux  in  the  tubular  element,  viz.,  Bldx. 
Therefore 

—  de=—  pldi  =  --  ^  -r.Bldx, 

J.  \J      Ct'L 

using  signs  of  partial  differentials,  and  re-arranging,  we  have : 

dx  =  W~p    dT 
Differentiating  (11)  with  respect  to  t  and  (5)  to  x, 

d2B       I  dB  di 

+  -    —  =  0.47T/*-  (13) 


(14) 


a*  ~  lov   at 

The  solution  of  equation  (15)  is  somewhat  difficult  and  is 
therefore  delayed  until  later  (see  Skin  Effect  in  Cylindrical 
Conductors). 

Equation  (16),  however,  which  shows  the  flux  distribution  in  a 
lamination  is  readily  solved  when  the  impressed  m.m.f.  and  there- 
fore, at  least  in  non-magnetic  materials  the  flux  density  is  a  sine 
function  of  time. 

Let  B  =  Bm  sin  cot 

—  - 

The  effective  value  of  the  first  expression  may 
be   represented  by  vector  OA  =  B,  and  that  of 

B      A    the  second  expression  in  the  derivative  by  OM  = 
FIG.  60. 


dxdt   '    x    dt  "  dt 

and  d2i          1     d2B 

dx2       108p  dxdt 

Substituting  (-12)  and  (14)  in  (13) 

d2i        \  di       0.4717*  di 
dx~2~^~  x  dx  =    108p    ~di 

For  long  or  thin  flat  bars, 

di 


Thus  dealing  now  with  effective  values  we  can  write: 


DISTRIBUTION  OF  FLUX  155 

where 


P108 

To  solve  this  equation  we  write, 

d2B 

—  -  v2B  =  0     /.  m2   -  v2  =  0 

.'.  m  =  ±  v 
and  B  =  A^vx  +  A^e~vx. 

Since  the  density  must  be  the  same  at  equal  distances  from  the 
center  line 

A^vx  -f  A2e~vx  =  Ai€~vx  +  AzeLvx 

which  requires  that  A  i  =  A2  =  A 

2  _  04^ 
plO8  ' 

It  is  readily  seen  that  if  v  =  (1  +  j)a 
V2  =  2ja2 


2j         P108 


-oa:~Jaz 


or  B  =  A[eaxejax  +  e-oa:c 

Substituting  trigonometric  expressions  and  combining  the 
real  and  imaginary  terms  we  get: 

Since  e±:iax  =  cos  ax  ±  j  sin  ax 

B  =  A[(eax  +  e~ax)  COS  ax  +  j(eax  —  eax)  sin  ax]. 

If  B  i  is  the  effective  value  of  the  surface  density  then  B  =  BI 
for  x  =  d. 

If,  furthermore,  it  is  assumed  as  an  approximation  that  e~a& 
is  small  compared  with  eaS  then, 

Bl  =  Aea5  (cos  ad  +  j  sin  a  8) 

and  flic"*' 

cos  a8  +  j  sin  ad 

t°*  +  e"°X)  COS  «  +     «"  -  «")  «n  ««] 


156  ELECTRICAL  ENGINEERING 


and  B  —  Bie~a  V  e2ax  +  e~2ax  +  2  cos  2  ax 

where  as  given  above 

<0.27r2/^7 

:  \   p io8 

For  iron  p  is  approximately  IO5  and  /*  may  be  anything  up  to 
18,000. 

1 


For  copper  p  is  approximately  „ 


and  i  =  1. 


PART  II.    PROBLEMS  IN  ELECTROSTATICS 

CHAPTER  XII 
FUNDAMENTAL  LAWS 

Coulomb's  Law.  —  The  fundamental  law  upon  which  our  know- 
ledge of  electrostatic  or  electromagnetic  phenomena  rests  was 
found  experimentally  by  COULOMB.  It  is  similar  to  NEWTON'S 
law  of  gravitation  and  is: 

F  =  c^orF  =  c^  (1) 

iv>i  ntZ 

where  F  is  the  force  acting  upon  the  point  charges  Qi  and  Q2,  or 
the  magnet  poles  of  strength  mi  and  w2,  c  is  a  constant  depending 
upon  the  system  of  units  employed,  and  r  is  the  distance  between 
the  charges  or  magnet-poles,  respectively. 

In  the  electrostatic  system  of  units,  as  well  as  in  the  electro- 
magnetic system  of  units,  c  is  taken  as  unity  when  the  medium 
is  air,  or  rather  vacuum, 

.:r-3&.,**r-*F-      .  (2) 

where  F  is  expressed  in  dynes. 

Thus  two  unit  charges  or  two  unit  magnet-poles  repel  each 
other  with  a  force  of  1  dyne  when  separated  1  cm. 

If  the  medium  has  a  specific  inductive  capacity  K,  then 

1 

= 


K        r* 
If  the  magnetic  medium  has  a  permeability  /z,  then 


The  strength  of  unit  poles  is  then  measured  —  assuming  that 
it  were  possible  —  by  the  repulsion  between  two  similar  poles. 
When  the  force  is  1  dyne  and  the  distance  is  1  cm.,  the  poles  have 
unit  strength. 

157 


158  ELECTRICAL  ENGINEERING 

Field  Intensity. — Surrounding  electric  charges  or  magnet-poles 
is  a  field,  and  the  intensity  of  the  field  at  a  point  is  defined  as  the 
property  of  the  space,  which  is  measured  by  the  force  exerted  by 
the  field  on  unit  charge  or  unit  pole  located  in  that  point,  when 
electric  and  magnetic  fields,  respectively,  are  considered. 

Because  of  this  definition,  it  must  not  be  inferred  that  the 
intensity  of  the  field  is  a  force;  it  is  not  a  force,  but  merely  a 
space  function  just  as  the  gravitational  field  intensity  is  a  space 
function.  The  force  acting  on  a  certain  mass  at  a  certain  point 
may  have  any  value,  depending  upon  the  particular  mass  used 
in  the  experiment. 

Important  Theorems. — While  COULOMB'S  law  forms  the  basis 
on  which  the  theories  rest,  the  progress  in  the  art  would  probably 
have  been  slow  were  it  not  that  a  number  of  theorems  have  been 
worked  out  more  or  less  directly  from  that  law.  These  theorems 
are: 

GAUSS'S  theorem,  the  divergence  theorem,  GREEN'S  and 
STORE'S  theorems,  etc.,  all  having  important  bearing  on  prac- 
tical problems. 

Surface  Integral  of  a  Distributed  Vector. — As  a  preliminary 
to  these  theorems  the  surface  integral  of  a  distributed  vector 
will  be  defined. 

It  will  be  assumed  that  an  electric  field  exists  due  to  some 
charge  and  that  lines  of  force  or  tubes  of  force  radiate  from  the 
charge  in  all  directions.  It  is  desired  to  find  the  number  of  lines 
that  go  through  a  surface,  say  a  cap  that  is  placed  in  the  field. 
In  Fig.  61,  AB  may  be  assumed  to  be,  for  instance,  the  inter- 
section of  the  plane  of  a  loop  of  wire,  over  which  the  cap  is  made, 
with  the  plane  of  the  paper. 

If  the  surface  of  the  cap  were  divided  up  into  a  number  of 
elements  and  the  direction  and  the  intensity  of  the  field  at  every 
point  were  known,  then  it  obviously  would  be  possible  to  calculate 
the  total  number  of  lines  (the  flux)  that  crosses  the  cap  or  the 
surface. 

The  sum  of  the  fluxes  normal  to  each  element  of  the  surface 
is  called  the  surface  integral  of  the  normal  field  intensity  over  the 
cap,  or  the  total  outward  flux  through  the  cap.  (The  normal  to 
the  elementary  surface  is  always  understood  to  be  drawn  out- 
ward from  the  surface.  On  account  of  the  sign  of  trigonometric 
function  a  normal  drawn  inward  will  lead  to  a  negative  surface 
integral.) 


FUNDAMENTAL  LAWS 


159 


If  another  diaphragm  or  cap  (Fig.  62)  were  stretched  across  the 
wire  loop  AB,  it  is  evident  that  a  certain  amount  of  flux  would 
enter  the  space  between  the  diaphragms  and  a  certain  amount 
leave  it. 

It  will  be  shown  in  this  case  that  the  total  normal  outward  flux 
from  the  space  enclosed  by  the  two  caps  will  be  zero,  as  long  as 
no  charged  particles  are  enclosed  by  the  diaphragms. 


Consider  then  a  distributed  vector  field  (Fig.  63),  and  let  R 
be  the  value  of  the  vector  at  the  small  surface  element  dS, 
making  an  angle  6  with  the  normal  to  the  surface  element.  R 
represents  the  electrostatic  or  electromagnetic  field  intensity. 

The  field  intensity  along  the  normal  is  then  R  cos  6,  and  the 
flux  going  through  dS  is:  dif/  =  R  cos  6dS,  i.e., 

df  =  R  cos  (NtR)dS. 

.'.  \l/  =  total  outward  flux  through  the  surface, 
RcosddS, 


where  N  is  the  component  of  the  field  intensity  normal  to  dS,  i.e., 
N  =  R  cos  0. 

This  can  also  be  expressed  in  rectangular  coordinates  by 
vector  analysis,  provided  that  dS  represents  not  a  surface  dS, 
but  a  vector  at  right  angles  to  dS  of  size  dS.  (See  appendix  for 
dot  product.) 

Let 


and 


R  =  X(x,y,z)i  +  Y(x,y,z)j  +  Z(x,y,z)k 
dS  =  dSxi  +  dSyj  +  dSzk. 


160  ELECTRICAL  ENGINEERING 

Then, 

ffR-dS  =  ff(XdSx+ 

~  ,~    '     ^,«N        /•  /»/T?J     j      where  obviously  dSz  =  dydz. 

YdSy  +  ZdS.)  =ff(Xdy-dz  _  y 


+  Zdxdy), 

Another  way  of  expressing  the  surface  integral  of  a  distributed 
vector  field  is: 

ff(Xl  +  Ym 


In  these  equations  X,  Y  andZ  are  the  components  of  the  vector 
along  the  three  axes  and  I,  m,  and  n,  the  direction  cosines  of  the 
normal  to  the  surface. 

Thus:  I  =  cos  (N,x), 

and,  m  =  cos  (N,y), 

n  =  cos  (N,z). 
.'.  IdS  =  dydz, 

and,  mdS  =  dzdx, 

ndS  =  dxdy. 

.'.  ff(XdSx  +  YdSy  +  ZdS.)  =  ff(Xl  +  Fw  +  Zn)dS. 

Gauss's  Theorem.  —  According  to  GAUSS'S  theorem  the  total 
normal  outward  flux  from  a  closed  surface  containing  a  charge 
Q  is  =  47rQ. 

Let  N  be  the  component  of  the  field  intensity  R  normal  to  an 
elemental  surface  of  the  bag  dS. 

The  theorem  can  be  expressed  mathematically  by: 

ffNdS  =-  47rQ. 

Let  dw,  Fig.  64,  be  the  solid  angle  at  A  corresponding  to  dS  or  dSi, 
which  is  perpendicular  to  R 


or,  dSi  =  r2w. 

ffNdS  =  SfRcosedS  =  ffRdSi  =  f  fRr*dvu. 
But  COULOMB'S  law  gives: 


_QQ 

or  since  by  definition 


F_i 

r2 


F  =  R, 


when 


FUNDAMENTAL  LAWS 
Qi  =  unity. 


161 


(1) 


FIG.  64. 

the  integral  to  be  taken  around  the  entire  surface,  that  is  over  the 
complete  solid  angle,  which  is  47r, 


Thus, 


=  47rQ. 

ffNdS  =  47TQ, 
or  if  there  are  many  charges  in  the  envelope, 

ffNdS  =  4irSQ. 

It  is  seen  that  the  total  flux  radiating  from  a  point  charge  Q  or 
a  magnet  pole  m  is  <p  =  4ir  Q  and  (p  =  4irm  respectively. 

It  will  be  shown  that  while  the  conception  of  lines  or  tubes  of 
force  is  very  much  the  same,  both  serve  to  map  out  a  field,  by 
convention  a  tube  includes  4?r  lines. 

From  (1)  it  is  seen,  that  the  intensity  at  a  point  distant  r  from 
a  point  charge  Q  is  —  2- 

By  a  similar  reasoning,  it  is  readily  seen  that  in  magnetic 
problems, 

tp  =  magnetic  flux  =  4?rm; 

and,  „       m 

H  -  r,, 

where  m  is  the  strength  of  the  pole  causing  the  field,  and  H  is  the 
intensity  of  the  magnetic  field. 


162 


ELECTRICAL  ENGINEERING 


But,  to  return  to  GAUSS'S  theorem,  it  is  readily  seen  that  the 
shape  of  the  bag  is  immaterial.  Assume  so,  for  instance,  that 
the  shape  is  that  shown  in  Fig.  65. 


FIG.  65. 


FIG.  66. 


The  vector  R  cuts  the  surface  three  times.  The  outward 
normal  flux  is  positive  at  A,  negative  at  B,  and  again  positive 
at  C.  Thus  the  net  result  is  one  positive  outward  flux  (Fig.  65). 
Were  the  charge  outside  of  the  envelope,  then  the  flux  cuts  the 
bag  two,  four,  six  or  an  even  number  of  times,  so  that  the  total 
outward  flux  is  cancelled  by  an  equal  total  inward  flux  (Fig.  66). 

The  net  result  then  is,  that 

ffNdS  =  0, 
when  the  bag  does  not  contain  a  charge. 

Potential.  —  The  electric  potential  is  similar  to  the  potential 
energy  of  matter;  it  is  a  space  function. 

The  electric  potential  at  a  point  is  defined  as  the  work  done 
in  bringing  a  unit  positive  charge  from  a  place  of  zero  field  to  the 
point  under  consideration. 

The  magnetic  potential  is  defined  in  a  similar  way,  substituting 
unit  pole  for  unit  charge. 


Path  of  Unit 
Charge 


FIG.  *67. 


Referring  to  Fig.  67  R  is  the  intensity  at  a  point  of  the  path  of 
the  unit  charge  in  its  journey  from  infinity,  where  the  field  is 
zero,  to  the  point  P,  where  the  potential  is  to  be  determined,  then 

V  =  -  f  R  cos  6ds, 


FUNDAMENTAL  LAWS  163 

the  minus  sign  being  adopted  by  convention;  but 

dr  =  ds  cos  6,  .'  .ds  =  -  —  , 
cosd 

and, 

rr  =  rp          Crpo        iirp    o 

V        -  Mr-     -I     -jUr  =  Q-M       .£. 

Jr   =    oo  J  oo       I  I    \    oo  '  p 

In  general,  the  potential  at  a  point  due  to  several  point  charges 


It  is  interesting  to  note  that  the  potential  is  not  dependent 
upon  the  path  chosen  in  the  journey;  'it  depends  only  upon  the 
point  charge  at  A  and  the  distance  between  P  and  A.  It  is 
strictly  a  space  function. 

The  potential  is  the  same  on  any  surface  the  elements  of  which 
have  the  same  distance  from  the  point  charge. 

Thus  the  potential  of  the  surface  of  a  sphere  having  a  point 
charge  in  its  center,  and  influenced  by  no  other  charge,  is: 


where  r  is  the  radius. 

Since  by  definition  the  capacity  is  C  =  y,  we  note  that  the 

capacity  of  an  isolated  sphere  is  C  =  r. 

The  capacity  in  the  electrostatic  system  of  units  is  in  centi- 
meters. A  sphere  of  10  cm.  radius  is  said  to  have  a  capacity 
of  10  cm. 

It  will  be  shown  later  that  to  convert  the  capacity  to  farads 

V2        (3  X  1010)2 
involves   a   division  by  ^  =  ~  ~Tnir~       =  ^  X  10  n.     Thus  in 

this  case  the  capacity  of  the  particular  sphere  would  be  C  = 
10       t 

~          farads- 


Line  Integral.  —  The  intensity  R  of  the  electric  field  has  not 
only  a  definite  numerical  value,  but  also  a  definite  direction. 

Let  the  components  of  R  along  any  three  rectangular  coordi- 
nates be,  X,  Y  and  Z,  and  let  the  components  of  the  distance  ds 

on  the  respective  axes  be  dx}  dy  and  dz.     Then,  since  the  poten- 
11 


164 


ELECTRICAL  ENGINEERING 


tial  is  the  same,  no  matter  what  path  we  may  take,  by  travel- 
ling along  the  axes,  we  get: 


fx,y,z 

V         J-     ' 


Xdx  +  Ydy  +  Zdz. 

This  integral  is  called  the  line  integral  of  the  distributed  vector 
along  the  path. 

Using  vector  analysis  (see  appendix)  we  get: 

V  =  —  J*R-ds,  the  integral  of  the  dot  product, 
where  R  =  iX  +  JY  +  kZ, 

and  ds  =  idx  +  jdy  +  kdz, 

.'.  R-ds  =  Xdx  +  Ydy  +  Zdz. 
Differentiating,  we  get: 

dV  =  -  (Xdx  +  Ydy  +  Zdz)  (1) 

Recollecting  that  if  V  is  a  function  of  the  space  coordinates  x,  y 
and  z  only, 


d 


dV 


It  is  evident,  since  it  has  been  shown  that  the  potential  is  a 
function  depending  only  upon  the  space  coordinates,  that 


~ 

do; 


and 


and 


ay 

dn 


=  -  z 


=  -  R 


(2) 


where  —  means  differentiation  along  the  lines  of  force,  —  T—  is 
dn  dn 

usually  denoted  by  G,  the  potential  gradient. 

Equation  (1)  must  be   a  complete   differential,   the  criterion 
of  which  is  that, 

ar  _  dx 

~dx    ~  dy  =      ' 


_ 
dy       dz 


FUNDAMENTAL  LAWS  165 

and  aX  ^  dZ^ 

dz  ~~  dx  ~~ 

dY  a27 

~dx  ~     ~  dydx 
and  dX  dzV 

dy  ~         ~dxdy 

Thus,  aF       dX 

— .—  =  0,  which  was  to  be  proven 

Gauss's  Theorem  in  Term  of  Potential  Gradient. — From  equa- 
tion (2)  it  is  evident  that  GAUSS'S  theorem  can  be  expressed  in 
yet  another  form. 

Since  R  =  —  -_ — .  it  is  evident  that 
dn} 

dS  is  the  total  outward  flux. 

^ 

Thus, 


if  the  envelope  contains  a  charge;  and 

-  dS  =  0,  if  the  envelope  contains  no  charge. 
dn 

In  both  cases,  T-  means  differentiating  along  the  normal  to 
on 

surface,  dS. 

On  account  of  the  similarity  between  the  electric  and  magnetic 
definitions,  we  obtain,  by  reasoning  identically  with  that  given 
above,  for  the  magnetic  potential, 

fx,y,z  fx,y,z 

V  =  -      I    H  cos  dds  =  -    I     (Ldx  +  Mdy  +  Ndz), 
where 

dV      *,  dV      Ar  dV  i   17  dV' 

L  =  —  — >    M  =  —   — »    N  =  —  -->    and  H  =  —  — » 

a^  a^/  a^  an 

and, 

7=2—;    I    J  —  dS  =  -  47rZra,    if    the   envelope    contains 

magnetic  particles;  and   I    I  — -  dS  =  0;    if  the  envelope  does 

J  J  on 
not  contain  magnetic  particles. 


166 


ELECTRICAL  ENGINEERING 


Equipotential  Surfaces  shown  in  Dotted  Lines  Around  Two  Point  Charges 

Separated  5  cm.;  Q  l  =  +1Q.;  Q2=  -  5. 

FlG.   68. 


Lines  of  Force  shown  in  Fine  Lines  and  Equipotential  Surfaces  shown  in  Heavy  Lines 
Around  Two  Point  Charges  Separated  5  cm.  ;  Q  t  =  + 10 ,  Q  2  =  -  5 

FIG.  69. 


FUNDAMENTAL  LAWS  167 

In  these  equations,  H  is  the  intensity  of  the  magnetic  field, 
L,  M  and  N  are  the  components  of  the  intensity,  H,  along  the 
three  axes. 

V 

Application  of  the  Formula,  V  =  2  -  — In  Fig.  68  is  shown  the 

equipotential  surfaces  between  two  point  charges,  Qi  =    +10  and 
Qz  =  —5,  separated  5  cm. 

The  potential  at  any  point,  P,  is  obviously : 

VP  -  «'  +  $. 

ri        r2 

The  lines  of  force  can  not  well  be  shown  in  a  plane,  but  a  fair 
idea  of  their  shape  can  be  gained  from  Fig.  69. 

The  direction  of  each  line  of  force  is  obtained  by  combining 
72 1,  the  intensity  at  a  point  due  to  Qi,  and  R2,  the  intensity  due 
to  Q2. 


CHAPTER  XIII 

METHOD  OF  IMAGES,  APPLIED  TO  THE  PROBLEM  OF 
POINT  CHARGES  +10  AND  -5,  SEPARATED  5  CM 

In  plotting  the  equipotential  surfaces  of  the  problem  given 
above,  it  is  readily  seen  and  proven  that  the  surface  of  zero 
potential,  Fig.  70,  is  a  sphere,  and  that  the  following  relations 
obtain: 


(1)     a  = 


FIG.  70. 


25 


_  50 

cm.  and  p  =  -     — 


Substituting,  we  get  a  =  =~ 
X  5  =  3.33  cm. 

It  is  evident  that  the  field  distribution  will  not  be  affected  if 
a  grounded  metallic  sphere  of  radius,  p,  at  a  distance  (D  +  a) 
from  the  positive  charge  is  surrounding  the  negative  point  charge 
at  B. 

And  it  is  also  evident  that  the  potential  will  be  the  same  ( =  0), 
if  the  charge  at  B  is  removed  altogether. 

168 


METHOD  OF  IMAGES 


169 


It  is  thus  evident,  that,  reversing  the  line  of  argument,  the 
potential  distribution  in  a  system  involving  a  point  charge  at  A 
and  a  grounded  sphere  of  radius  p  with  center  at  a  distance  L  from 
the  point  charge,  can  be  determined  without  the  laborious  deter- 
mination of  the  distribution  of  the  induced  charge  on  the  sphere, 
simply  by  using  two  point  charges  at  A  and  B. 

The  location  of  B  and  the  charge  which  must  be  assumed  at  the 
non-existing  point  B  can  be  determined  from  the  following  rela- 
tions which  are  easily  proven: 

LP(L  -  D)  =  P2,  or  D  =  L  -  p*> 

L 

and  ~   p 


Potential  Distribution  between  a  Point  Charge  and  a  Metallic 
Sphere. — While  it  is  evident  then  that  the  field  can  be  determined 
without  much  labor  in  the  case  of  a  grounded  sphere,  the  problem 
becomes  quite  involved  if  the  sphere  is  insulated  and  kept  at  a 
certain  potential,  V,  which  is  not  zero. 

To  calculate  the  potential  distribution  in  that  case,  it  is  neces- 
sary to  study  the  distribution  of  the  surface  charge. 


FIG.  71. 

Consider  first  the  case  of  the  grounded  sphere.  The  intensity 
of  the  field  at  a  point  is  the  resultant  of  the  intensities  due  to  the 
charges  at  A  and  5,  the  so-called  inverse  points. 

It  must  be  expected  that  the  direction  of  the  resultant  field 
is  perpendicular  to  the  surface  at  all  points,  thus  we  can  draw  the 


170  ELECTRICAL  ENGINEERING 

diagram   in   Fig.    71.     Remembering   that   R}   the   intensity   is 
the  vectorial  sum  of 


-4  and  — v  or  of  Ri  and 


ri2          r2 


The  intensities  R\  and  R2  are  resolved  along  the  radius  CP  ana 
along  a  line  parallel  to  AB.  It  is  seen  that  PE  and  PG  are  equal, 
and  cancel  each  other,  so  that  the  resultant  intensity  is  in  line  of 
CP  and  is  PF  +  PH,  algebraically. 

Let  the  radius  of  the  sphere  in  Fig.  71  be  p.  By  similar  tri- 
angles, 

PF      PC  PC  p. 

~R~    =  PA    '  =/*/!•  p^  :  =  *ti  •  — , 

and  PH  _  PC        _T        „     PC 

Po         "    P#'    " 


But 
and 


And  it  has  been  shown  that  Q2  =  —  Qi  j-' 

Li 


Since  La  =  p2,  from  the  figure,  it  is  seen  that, 
n       L  p 


'     C    2  72\  C1\ 

p7?(p  -L) 

By  Coulomb's  theorem,  47r(T  =  .R,  where  o-  is  the  surface  density 
of  charge,  or  charge  per  square  centimeter, 

-£2)  (2) 


METHOD.  OF  IMAGES 


171 


When  the  radius  is  very  large,  the  surface  of  the  sphere  ap- 
proaches a  plane,  Fig.  72,  and  a  approaches  p.  Thus,  if  d,  in 
Fig.  72,  is  the  distance  of  the  point  charge  from  the  plane  of 
zero  potential,  we  have: 

L  =  P  +  d, 
which,  substituted  in  (1),  gives: 


R  =     3  (p2  ~  (p 


=:      (~  2pd  ~ 


or,  since  d-  is  small  compared  with  2pd, 


and  the  surface  density  of  charge  is: 

Qid 

27rri3 


FIG.  72. 


The  surface  density  of  charge  decreases  inversely  as  the  cube 
of  the  distance  from  the  point. 

Assume  now  that  the  sphere  is  insulated  and  without  charge, 
it  will  then  have  some  potential  not  zero. 

It  was  shown,  that,  when  the  sphere  is  at  zero  potential,  it 

acts  as  if  it  had  a  charge  Qz  =  —  Q\Y  ^  ^ne  inverse  point  B 

of  point  A.  In  order  that  its  charge  shall  be  zero,  we  have  to 
apply  mathematically,  somewhere  in  the  sphere,  a  charge  = 

—  Q2  =  _|_Qiy.     Then  the  total  charge  obviously  is  zero. 
Li 

Since  the  resultant  potential  of  the  external  charge  Qi  and  the 
internal  charge  —Q\j  gives  zero  potential  of  the  spherical 

surface,  in  order  to  maintain  a  uniform  potential  V  all  over  the 
sphere,  the  assumed  charge  must  be  applied  in  the  center  of  the 
sphere. 

Thus  we  deal  with  three  charges,  which  combined  cause  the 
external  field. 

First.  —  The  field  due  to  the  external  point  charge  Q\. 

Second.  —  The  field  due  to  the  charge  Q2  at  tne  inverse  point. 

Third.  —  The  field  due  to  the  charge  —  Q2  in  the  center  of  the 
sphere. 


172  ELECTRICAL  ENGINEERING 

The  charge  —  Q2  gives  a  uniform  surface  density  of 

Q2  (?i 

47TP2  *~   47TPL 

The  combined  effect  of  Qi  at  A  and  Q2  at  5  has  been  shown  to 
give  a  surface  density  of 


= 


Thus  the  actual  surface  density  is: 


Equation  (3)  then  gives  the  distribution  of  the  surface  charge 
on  an  insulated  sphere  without  any  independent  charge.  The 
equation  must,  and  does  show,  that  a  is  positive  on  one  side  and 
negative  on  the  other  side,  in  order  that  the  total  charge  be  zero. 

The  potential  of  the  sphere  is  obviously, 

charge  _  Qip  _  Qi 
radius        pL        L 

This  is  of  interest,  in  that  it  shows  that  the  potential  of  a  sphere 
due  to  a  point  charge  Qi  situated  L  cm.  from  the  center  is  -j^- 

This  can  be  proven  in  a  more  general  way  as  follows  : 

Assume  that  a  non-conducting  sphere  be  placed  in  an  electric 
field  caused  by  a  number  of  point  charges,  a,  6,  c,  etc.     Let  the 
potential  of  a  small  element  of  the  sphere  be  V.     The  value  of 
V  changes  from  point  to  point  of  the  surface  of  the  sphere. 
The  average  value  of  the  potential  V  is: 

VdS 


where  dS  is  an  element  of  the  surface. 
Referring  to  Fig.  73  : 

dS  =  r  sin  6d4>rdd 

.'.  Vm  ^^ffVs 
and  the  average  potential  gradient  along  the  radius  is: 


*       Cf 

irr2  J  J 


dS. 

4irr2  J  J    dr 

Since  —  is  the  intensity  as  well  as  the  gradient  it  follows  that 


METHOD  OF  IMAGES 


173 


—  dS  is  the  flux  diverging  from  the  sphere.     This  is  zero  as  we 
have  assumed  that  no  charge  exists  in  the  sphere. 

Thus  -~^  —  0  and  Vm  is  a  constant  for  all  values  of  r. 

We  conclude  then  that  the  average  potential  of  a  sphere  is 
the  same  as  the  potential  at  the  center. 


rdO 


FIG.  73. 

Suppose  now  that  the  insulated  sphere  had  a  charge  QQ. 

In  order  that  the  surface  of  the  sphere  shall  be  an  equipotential 
surface,  this  charge  also  should  be  considered  as  placed  in  the 
center,  and  its  surface  density  should  be  added  to  those  given 
above. 

L  n3 


The  potential  of  the  sphere  will  obviously  be  the  sum  of  the 
potentials  due  to  its  own  charge  Q0  and  due  to  the  point  charge 

Qo   ,   Qi 


or, 


Usually  V  is  known  rather  than 


It  is  of  interest  to  find  the  attractive  or  repulsive  force  between 
the  point  charge  at  A  and  the  sphere. 


174  ELECTRICAL  ENGINEERING 

The  force  is,  by  COULOMB'S  law,  proportional  to  the  product 
of  charges  and  inversely  proportional  to  the  square  of  the 
distance. 

The  following  conditions  therefore  exist: 

First. — A  charge  +  Qi  at  A.     Fig.  74. 

Second. — A  charge  —  Q\  y-  at  B. 

Li 

Thisd.—A  charge  +  #1  ~  +  Q0  at  C. 
Li 


-<                       D 

^  r  

V  -  -  : 

FIG.  74. 
Thus  the  force  between  the  sphere  and  the  point  charge  is: 

F  = 


L2  —  p2 

But  it  has  been  shown  that  D  =  — j > 

Lt 

p  =    ~  Qi2L2     jo      Qi2p      QoQi 
"  (L2-p2)2"  L~*    L3          L2  ' 

which,  by  transformations,  becomes : 

„       QoQi      0,p3(2L2-p2) 
L2      "  Vl  L3  (L2  -  p2)2 
Example— Qi  =  1,  QQ  =  10. 
L  =  variable, 
p  =  10. 
For  L  =  100, 

10       1000     20,000  -  100  1 

104    '    106    (10,000  -  100) 2  "      *  1000    yn 
For  L  =  11, 

10        1000  (242  -  100) 
F    -  121-  1330  (121  -10Q)2=     -  0.158  dyne  attraction. 

It  is  thus  seen  that  a  lightly  charged  particle  may  be  repelled,  if 
far  away  from  a  charge  of  the  same  sign,  and  may  be  attracted 
when  near.  If,  however,  the  charges  are  of  opposite  signs,  the 
charges  attract  always. 


METHOD  OF  IMAGES 


175 


Problem. — Construct  the  equipotential  surfaces  between  an 
insulated  charged  sphere  and  a  point  charge,  when 

p  =  10  cm., 
L  =  20  cm., 

Qi  --  i, 

and  _  21 

~  20' 

Potential  Distribution  between  Two  Spheres. — Let  sphere  A 
in  Fig.  75  have  a  pot.  V  and  a  radius  R]  and  sphere  5  have  a 
pot.  FI  and  a  radius 


FIG.  75. 

Calculate  first  the  charges  at  A  and  B  and  the  location  of 
these  charges,  when  A  is  at  potential  V  and  B  is  at  zero  potential. 
Then  reverse  the  operation,  and  calculate  the  charges  at  A  and  B 
and  the  location  of  these  charges,  when  B  is  at  a  potential  Vi 
and  A  is  at  zero  potential.  Then  add  the  charges  and  potentials 
respectively,  and  the  desired  solution  is  obviously  obtained. 

(1)  Calculation  of  the  charges  on  A  and  B  when  the  potential 
of  A  is  V  and  that  of  B  is  zero : 

The  first  approximation  is  obtained  when  the  potential  of  A 
alone  is  considered.  We  have  then,  since  in  general  Q  =  VR,  a 
charge  in  the  center  of  A  of  value  Q0  =  VR,  and  we  may,  for 
completeness,  say  that  its  distance  a0  from  the  center  is  zero. 

VR 
This  charge  affects  B  by  giving  B  a  potential,  which  is  -y— 

Since,  however,  the  potential  of  B  must  be  zero,  it  is  necessary 

VR 

to  supply  B  with  a  charge  which  gives  a  potential  -    -j-.     This 

charge,  which  may  be  called  Q'i>  has  previously  been  shown 

/7?  \ 
to  be  Q'i  =  —  VR(-J-},  shall  not   be  placed  in    the    center  of 


176  ELECTRICAL  ENGINEERING 

the  sphere,  but  at  a  distance  61,  which  is  obtained  by  the  relation 
previously  proven: 

(radius)2 

~  distance  from  charge  to  center  of  sphere 


\T~D~D 

But  the  charge  Q'i  or  --  j  —  at  61  affects  the  potential  of  A, 
so  that  its  potential  is  no  longer  V,  but 
V  +     f  --  r  —  -j  -s-  (distance  from  charge  Q'i  to  center  of  A) 

V-       VRRl 

L(L  -  &0* 

To  bring  the  potential  of  A  back  to  V,  A  must  be  supplied  with  a 
charge,  which  is: 


As  far  as  the  external  action  of  the  charge  is  concerned,  it  is 
located  at  ai}  where  as  before 


This  charge  at  a\  affects  sphere  B  and  induces  a  potential  which  is 

VR2Ri 

L(L-bi)     (L-ai)' 
In  order  to  bring  the  potential  back  to  zero,  a  charge  Q'%  has  to 

_    T/P2P 

be  added  to  B,  which  gives  a  potential  of  r  /r  _  ^  \  —  (T~I  —  )' 

and  this  charge,  as  far  as  external  influences  are  concerned,  is 
located  at  a  point  62,  where 

,  RS 

62  = 


Continuing  the  process,  the  necessary  additional  charge  on  A  to 
balance  the  effect  of  Q'%  at  62  is  found  to  be : 

Qi=_^_^_  = 

and 

Again,      n, 

v  3       L(L  -  61)  (L  -  ai)  (L  -  62)  (L  -  at)' 
and  ,  R\2 

°3  "  (L-a2)' 


METHOD  OF  IMAGES 


177 


The  total  charge  on  A  is  QA  =  Q0  +  Ql  +  Q2  +  .    .    .    ; 

The  total  charge  on  B  is  QB  =  Q'i  Q'z  +  Q'z  +  .    .    .   . 

But,  it  must  be  remembered  that  in  order  to  find  the  intensity 

of  the  field  at  any  point,  the  position  of  the  charges  has  to  be 

considered. 

,  (2)  By  an  identical  method,  a  new  set  of  charges  are  obtained, 
when  A  is  kept  at  zero  potential  and  B  at  its  potential  V. 

The  total  charges  on  A  and  B  are  the  sum  of  all  the  charges  so 
calculated. 

Assuming,  for  instance,  that  the  potentials  of  A  and  B  are 
both  positive. 

The  first  set  of  calculations  will  then  give  a  number  of  positive 
charges  in  A,  all  of  which,  except  the  first,  located  at  points,  riot 
its  center,  the  charges  in  B  will  all  be  negative,  and  all  be  located 
at  points  not  its  center. 

The  second  set  of  calculations  (not  shown  above)  will  result  in 
a  series  of  negative  charges  in  A,  all  of  which  are  located  at  points 
not  its  center,  and  a  set  of  positive  charges  in  the  sphere  B}  the 
first  of  which  is  at  its  center.  Thus  the  total  charge  in  either  A 
or  B  is  a  sum  of  a  series  of  positive  and  negative  charges. 

Simple  Case. — For  two  similar  spheres,  one  at  zero  potential 
and  the  other  at  a  potential,  V,  we  have: 


On  the  sphere  of  pot.  V 

On  the  sphere  of  pot.  zero 

Q0  =  VR 

a0    =  0 

o'          VR2 
Ql  =      T 

h     R2 

bl==~L 

o         VR3 

o'                  VRi 

L(L  -  6x) 
R2 

L(L  -  bJ(L  -  ax) 

t              ^ 

ai~    (L  -  60 

(L  -  at) 

Q                              VR* 

Q's  - 
VR6 

L(L  -  6X)(L  -  ai)(L  -  62) 
R2 

L(L-61)(L-a1)(L-62)(L-a2) 
b>    -          RZ 

°*        (L  -  62) 

(L  -  a,) 

178  ELECTRICAL  ENGINEERING 

The  total  charge  on  the  sphere  of  potential  V  is : 

QA  =  Qo  +  Qi  +  Q2  +  .   .    .   ; 

and  that  of  the  sphere  of  zero  potential  is : 

QB  -  Q'l  +  Q'2  +  Q'z  +  .    . 

To  study  the  sphere  gap,  the  following  problem  has  been 
solved  to  show  more  particularly,  that,  while  the  difference  in 
potential  between  two  gaps  may  be  the  same,  one  gap  may  break 
down  with  considerably  lower  potential  difference  than  the 
other. 

Air  at  atmospheric  pressure  appears  to  sustain,  as  a  maxi- 
mum, a  density  of  about  100  lines  per  sq.  cm.,  or  a  potential 
gradient  of  100,  electrostatic  units  or  in  practical  units  30,000 
volts  per  cm.  If,  therefore,  the  potential  to  ground  is  high,  the 
air  may  well  break  down  around  the  spheres,  even  though  the 
potential  difference  between  the  spheres  may  be  comparatively 
low. 

When  the  air  breaks  down,  corona  appears.  Then  the  effective 
dimensions  of  the  spheres  are  increased  and  the  gap  length 
correspondingly  lowered. 

The  following  three  cases  are  calculated,  and  the  results  are 
tabulated  below. 

Diameter  of  the  spheres,  25  cm. 

Distance  between  surfaces,  14  cm. 

Potential  difference  1000  electro  static  units  or  300,000  volts. 

In  the  first  case,  sphere  A  has  a  potential  of  1000  and  B  is 
at  zero  potential,  in  the  second  case  the  spheres  are  at  potentials 
+500  and  —500  respectively,  and  in  the  third  case  they  are 
at  potentials  +1500  and  +500  respectively.  In  the  example 
the  potential  gradient  G  is  calculated  at  the  surface  of  the  sphere 
of  highest  potential  on  the  center  line  between  the  spheres  al- 
though it  may,  of  course,  be  greater  at  some  other  points.  In 

general  G  =  —  S  .,.     .2-     The  gradients  due  to  the  two  spheres 

should  obviously  be  added  if  the  charges  are  of  opposite  potential. 
Since  the  intensity  of  the  field  is  in  the  same  direction  at  the  point 
considered. 


METHOD  OF  IMAGES  179 

Summary  of  the  first  case: 

For  sphere  A, 

a0  =        0  Co  =  12,500 

ttl  =  4.46  Ci  =     1,430 

a2=4.53  Q2  =        186 

a3=4.53  Q3  =      24.4 

For  sphere  B, 

60  =  0  Q'o  =  0 

bl  =  4  Q'i  =  -  4,000 

62  =  4.5  Q'2  =  -       516 

63  =  4.53  Q'3  =  --   67.3 

64  =  4.53  Q\  =  -           9 

In  general  G  =  ^ 

.'.G  =  -  S  ~  =      -  114.6,  or,  -  34,500  volts  per  cm. 

Thus  the  sphere  probably  begins  to  glow. 

Summary  of  the  second  case : 

ao  =  60  =        o  Qo  =  Q'o  =  6,250 

Ol  =  bi  =  4.01  Qi  =  Q'i  =  2,000 

a2  =  62  =  4.45  62  =  Q'2  =     714 

a3  =  63  =  4.51  Q3  =  Q's  =     258 

a4  =  &4  =  4.53  Q4  =  Q\  =   93.5 

a5  =  66  =  4.54  Q6  =  Q'6  =    34.2 

(^  =  -100.2  or  about    -30,000  volts  per  cm.     The  spheres 
ought  to  be  just  about  on  the  point  of  glowing. 

Summary  of  third  case : 

Qo  =        18,750  Q'o  =  6,250 

Ql  =  -    2,000  Q'I  =  -  6,000 

Q2  =  +    2,140  Q't  =  +     714 

Qs  =  -       258  Q;3  =  -     775 

#4  =  +       280  e;4  =  +    93.5 

Qb  =  -        34.2  Q'B  =  -  102.6 

The  a's  and  6's  are  the  same  as  above. 

G  =  -128  or  -38,400  volts  per  cm. 

Thus  the  spheres  glow  undoubtedly,  and  if  "ground"  is  under 
the  spheres  the  potential  gradient  may  be  slightly  higher  below 

the  line  connecting  the  centers  of  the  spheres. 
12 


CHAPTER  XIV 

APPLICATION  OF  THE  POTENTIAL  FORMULA  V  =  2  - 
TO  SOME  MAGNETIC  PROBLEMS 

The  magnetic  potential  at  a  point  in  a  magnetic  field  is,  as 
has  already  been  stated,  the  work  done  in  ergs  in  bringing  a  unit 
pole  from  infinity,  or  a  point  of  no  magnetic  field,  to  the  point 
under  consideration. 

By  GAUSS'S  theorem  the  outward  normal  flux  from  a  pole  of 
strength  m  is  4irm.  Thus  the  intensity  of  the  magnetic  field,  H, 

at  a  distance,  r.  from  the  pole  is  | — ^  '•> 

or,  H  =  m-f 

r2 

tlll(l)  .  -|-T  flv         7  "V. 


or  in  general,  V  =  2  —  • 

Obviously,  a  magnetic  pole  can  not  exist  alone;  there  is  always 
a  north  pole  and  a  south  pole  in  every  magnet.  Thus  to  get 
the  potential  at  a  point,  at  least  two  poles  of  opposite  signs  must 
be  considered. 

The  potential  of  a  small  magnet  at  distance  large  compared 
with  its  dimension  is: 

V  =  -  '  °°S    >  where  6  is  the  angle  the  axis  of  the  magnet  makes 

with  the  radius  vector  to  the  point. 

This  is  readily  seen,  if  the  magnetism  be  assumed  as  con- 
centrated at  the  poles  of  the  magnet. 
Referring  to  Fig.  76,  the  potential  at  P  is: 

_    m  m m 

V  =  AP  +    ~BP~ 


m  ,        —         (1) 


180 


THE  POTENTIAL  FORMULA  181 

If  r  is  large,  compared  with  I  then  V  = 


m  m 


A/r2  +  lr  cos  0       \/r2  —  Ir  cos  0 

The  square  root  can  be  expanded  by  the  binomial  theorem. 
We  have, 


=  —  |l  --  J^  -  cos  0  -h  .    .    .    .'  —  1  —  J£  -  cos  0  .    .    .1 

= %  cos  6  (approximately)  (2) 


Aba. 

FIG.  76. 

It  is  seen  from  (1)  that  the  magnetic  potential  at  P  is  in  times 
the  difference  in  -,  as  we  go  from  one  pole  to  another,  where  r  is 
the  distance  from  a  pole  to  the  point  P.  Let  I  =  ds,  then  the  rate 
of  change  of  -  along  ds,  is: 

-   (-)  .  thus  the  total  difference  is  —   (-)ds, 
ds  W  ds  W 


. 

dr 

If  I',  m',  and  nf  are  the  direction  cosines  of  the  magnetic  particle 
at  (x,  y,  z),  we  can  then  also  write, 

>""[>•  I  ©+»'sC')+»i  OK 

Magnetic  Shell.  —  A  thin  piece  magnetized  at  right  angles  to  its 
surface  is  called  a  magnetic  shell.     It  can  thus  be  assumed  as 


182 


ELECTRICAL  ENGINEERING 


made  up  of  a  large  number  of  small  magnets  as  shown  in  Fig.  77. 
Let  the  total  pole  strength  in  Fig.  78  be  m  and  the  area  S,  then  the 

pole  strength  per  unit  area  is  -«.     Let  the  thickness  of  the  shell  be 
I,  then  the  potential  at  P  due  to  the  shell  is  from  equation  (2). 


v 


S 


where  «  is  the  solid  angle  at  P  subtended  by  the  surface  of  the 
shell. 

(Recollect  that  the  solid  angle,  doi  =  -y  cos  0.) 


FIG.  77. 


FIG.  78. 


ml  is  called  the  magnetic  moment,  the  strength   of  a  mag- 
netic shell  or  the  moment  per  unit  area  is  usually  denoted  by  0, 

ml 


and,  V  =  grco. 

WEBER  proved  experimentally  that  a  small  circuit  in  a  plane 
carrying  current  produces  the  same  kind  of  a  field  as  a  magnet, 
and  that  the  potential  at  a  point  depends  upon  the  area  A  of  the 
coil,  the  current  7,  and  the  distance  to  the  point,  by  a  relation : 
_.       KAI  cos  0 

V        =         y > 

r2 

from  which  the  electromagnetic  unit  of  current  can  be  deter- 
mined by  making  k  unity. 

AI  cos  0      ml  cos  0 


thus,  and  7  =  —r-  =  g,  the  strength  of  the  magnetic  shell  sur- 
rounded by  the  circuit  or  coil. 


THE  POTENTIAL  FORMULA 


183 


Since  we  have  proven  that  V  =  #co,  we  get  the  following  simple 
relation  between  magnetic  potential  and  current: 

V  =  7co,  where  co  is  the  solid  angle  subtended  at  the  point  by 
the  surface  of  the  coil.  It  is  evident  then,  that,  as  long  as  we  do 
not  tread  the  circuit,  and  as  long  as  we  return  to  the  starting 
point,  the  work  done  in  moving  a  pole  in  the  field  is  zero. 

To  illustrate  this,  the  potential  at  a  point  on  the  axis  of  a  cir- 
cular wire  carrying  I  abs.  amp.  will  be  determined. 

First  let  the  point  be  at  the  center  of  the  coil,  Fig.  79,  then  co  = 
27r,  and,  V  =  2?r/. 


FIG.  79. 


FIG.  80. 


If  the  point  is  on  the  axis,  but  a  distance  x  from  the  face  of  the 
coil  as  shown  in  Fig.  80,  then  the  solid  angle  is: 


co  =  2ir(l  -  COS  a)   =  27r(  1 
and,  -  /  x 


'  The  magnetic  field  intensity  along  the  z-axis,  which  is  the 
direction  of  the  magnetic  field,  is: 


dx  ~  (R2  +  x*)*  ' 
and  the  force  in  dynes  on  a  pole  of  strength  m  is : 

2TrR2m 

=   (R2  +  X*)*  ' 

for  x  =  0,  that  is  if  the  point  is  in  the  plane  of  the  coil  and  in  its 
center, 


H  = 


27T/ 

R  ' 


The  work  done  in  bringing  unit  pole  once  or  several  times 
through  a  loop  carrying  a  current  /  will  now  be  investigated. 


184  ELECTRICAL  ENGINEERING 

Referring  to  Fig.  81,  before  the  journey  starts,  the  potential  at 
P  has  been  shown  to  be  7<o. 

When  the  journey  has  covered  1  revolution,  the  solid  angle 
has  changed  from  0  to  4?r.  Thus,  after  n  revolutions  of  the  unit 
pole  the  potential  of  it  is : 

Iw  +  4irln  =  I  (co  +  47rn). 

It  is  evident  then,  that,  when  a  magnetic-pole  of  strength  m  is 
moved  around  in  a  field,  and  returned  to  the  starting  point,  work 
will  be  done  every  time  the  circuit  is  treaded.  If  it  is  treaded  n 
times,  the  work  is: 


The  magnetic  potential  is  thus  a  multi-valued  function  of  the 
space  coordinates. 
p 


Path  of  Unit  Pole 

FIG.  81.  FIG.  82. 

dV 

The  intensity  of  the  magnetic  field  at  the  point,  H  =  —  -^ 

depends,  however,  only  upon  the  term  involving  the  solid  angle 
co,  not  upon  the  term  involving  4?m. 

Consider  now  a  straight  infinitely  long  wire  carrying  current  I. 

Let  the  wire  form  the  y-axis  and  let  the  point  be  in  the  x-z 
plane  (Fig.  82).  The  cone  subtended  by  the  plane  of  the  current 
(x-y  plane  with  ?/-axis  as  one  edge)  which  goes  out  to  infinity  and 
the  point  P  has  a  solid  angle,  2(ir  —  0). 

NOTE. — If  the  angle  in  the  x-z  plane  had  been  TT,  the  solid 
angle  would  have  been  27r;  in  this.case  the  former  is  (TT  —  0),  the 
latter  is  2(ir  -  0). 

.*.  V  =  /(«  +  4arri)  =  (27r  -  20  +  47m). 

The  direction  of  the  lines  of  force  which  are  circles  around  the 
y  axis  are  along  the  arc,  rdB,  then 

ff  =  ~S  =  T' 

an  equation  very  often  used  in  electrical  engineering. 


CHAPTER  XV 


R  (read  del  dot  R}, 
V  is  sometimes  called  LAME'S 


DIVERGENCE  OF  A  VECTOR,  POISSONS  AND  LAPLACE 

EQUATION 

It  has  been  shown  by  GAUSS'S  theorem  that  the  total  flux 
entering  and  leaving  a  closed  surface  in  a  vector  field  is  zero, 
unless  the  (closed)  surf  ace,  contains  some  charge  Q,  in  which  case 
the  outward  flux  equals  4?rQ. 

This  charge  may  be  a  single  charge,  or  it  may  consist  of  a  large 
number  of  small  charges  throughout  the  interior  of  the  surface. 

The  divergence  of  a  vector  is  the  excess  of  outgoing  flux  over 
the  incoming  flux  per  unit  volume  of  the  space  enclosed  by  the 
surface;  it  is  the  number  of  lines  which  diverge  per  unit  volume. 

If  the  excess  of  flux  in  a  small  volume  dv  is  d\f/,  then  the  diver- 

d\L> 
gence  of  the  vector  is  -T-» 

It  is  written  div.  R,  div.  (X,  Y,  Z)  or  V 

where  V  stands  f or  —  +  ^r  +  ^ 
dx       dy       dz 

differential  parameter. 

It  is  evident,  from  what  has 
been  said  above,  that  unless 
some  charges  are  enclosed  in 
the  small  volume,  there  can  be 
•no  divergence.  If  there  are  as 
many  units  of  positive  charge 
as  of  negative  charge  in  each 
small  volume,  there  can  also  be 
no  divergence,  i.e.,  div.  R  =  0. 
The  divergence  is  positive,  if 
there  is  an  excess  of  positive 

charge;  it  is  negative  (sometimes  called  convergence),  if  there  is 
an  excess  of  negative  charge.  The  presence  of  divergences 
involves  the  presence  of  charges.  In  hydraulics  the  presence  of 
divergence  means  either  the  presence  of  some  source  of  fluid  in 
the  element  or  some  change  in  density. 

Consider  a  small  volume  represented  by  a  cube,  in  Fig.  83  for 
the  sake  of  simplicity.  This  cube  is  assumed  to  be  a  small  part 

185 


T  Axis 


FIG.  83. 


186  ELECTRICAL  ENGINEERING 

of  the  total  volume  enclosed  by  the  envelope  that  contains  the 
charges. 

Let  X,  Y  and  Z  be  the  components  of  the  field  intensity  R 
parallel  to  the  coordinate  axes  and  at  the  center  of  the  surface 
a,  b  and  c. 

If  R  is  a  continuous  function,  which  depends  upon  the  space 
coordinates  only,  and  if  the  edges  of  the  cube  are  dx,  dy,  dz  then 
the  value  of  the  ^-component  of  the  field  intensity  at  Ci  =  Zi  = 


Thus,  the  incoming  flux  at  c  is:  Zdxdy,  the  outgoing  flux  at 
is  [Z  +  -T-  dz\  dxdy. 

Consequently,  the  difference  is 
'dZ 


(—dz}dxdy; 
Similarly,  for  the  other  sides, 


and  / 


-  dy }  dzdx. 


dy 

The  total  diverging  flux  is  thus: 


Hence  by  definition 

div.  R  =  V  -  R~  =  V-R. 

dv 

If  p  is  the  charge  per  unit  volume  or  the  volume  density,  then 
the  outward  normal  flux  is  4?rp. 

ax     ar     az 

dx         dy        dz' 

A  vector  field  is  said  to  be  solenoidal,  if  there  is  no  divergence. 
Such  a  field  is,  for  instance,  the  electric  field  in  free  space  or  the 
field  of  force  of  gravitation  in  free  space. 

The  divergence  theorem  connects  the  surface  and  volume  inte- 
grals and  states  that  the  surface  integral  of  the  normal  outward 
flux  of  a  distributed  vector  is  equal  to  the  volume  integral  of  the 
divergence  taken  throughout  the  volume.  It  is  one  of  the  forms 
of  Green's  theorem. 


ELECTRICAL  ENGINEERING  187 

It  is 

ffR  cos  SdS  = 


Using  the  notation  of  vector  analysis,  we  get: 
ffn-RdS  =  fffRdv, 

where  n  is  the  unit  normal  vector. 

This  theorem  is  subject  to  rigid  mathematical  proof,  but  can  be 
understood  without  advanced  mathematics,  if  the  volume 
enclosed  by  the  surface  is  assumed  to  be  divided  up  into  a  large 
number  of  small  volumes,  each  fitting  tightly  against  the  others. 

As  we  add  the  normal  outward  fluxes  of  the  different  elemental 
volumes,  all  will  cancel,  except  those  on  the  very  outside  surface, 
since  every  wall  separating  two  elements  is  integrated  over  twice 
with  normals  in  opposite  direction. 

The  outward  normal  flux  is  J*  J*  R  cos  Ods.  Since  the  excess 
of  outgoing  flux  over  the  incoming  flux  in  the  element  of  volume, 
dxdydz,  is: 

«  --  h  ~^~  +  -Q-J  dxdydz,  it  follows  that  the  total  outgoing 
flux  is: 

III    f-r—  +  -.J  —  h  -Q-}  dxdydz,    which     is    equal   to    I     I  R 

cos  6dS. 

Poisson's  equation  is: 

d*V\ 

"       =       "p- 


This  becomes:          dX       dY       dZ 


If  X}   Y  and  Z  are  gradients  or  intensities  of  a  scalar  point 
function  V,  so  that 

X  =  -  -V-  Y  =    -  d~-  Z  =    -  — • 

d#  '  dw '  62 


ax 

ay 

-ay  = 

az 

62  =          dz2' 


188  ELECTRICAL  ENGINEERING 

and 


dx*  "  dy2  "  dz*  =         47rp> 

where  p  is  the  density  of  electrification  or  charge  per  unit  volume. 

This  equation  then  applies,  when  the  region  of  the  electrostatic 
field  under  consideration  contains  positive  or  negative  charges,  or 
sources  and  sinks  as  some  writers  call  them. 

Laplace's  equation  is: 

VV       VV       d»F  _ 

dx*     ~  dy2  "  dz2   ' 

or,  as  it  is  often  written, 

V2F  =  0, 

(Read  del  square  V  =  zero)  and  refers  to  a  region  in  which  there 
are  no  charges,  or  to  a  solenoidal  field. 

By  means  of  LAPLACE'S  equation  it  is  possible  to  determine  the 
potential  at  any  point  in  the  dielectric  surrounding  a  charged 
body.  If  the  body  is  unsymmetrical  in  every  way  the  equation 
becomes  very  involved,  but  if,  as  is  almost  always  the  case  in 
practice,  there  is  some  axis  of  symmetry  and  particularly  if  the 
body  has  circular  symmetry  then  the  potential  distribution  can 
usually  be  calculated  fairly  ,  easily,  especially  if  a  table  of 
LEGENDRE'S  coefficients  is  available. 


CHAPTER  XVI 


LEGENDRE'S  FUNCTION 

The  potential  at  points  outside  of  the  bodies  having  circular 
symmetry,  such  as  circular  discs,  circular  rings,  etc.,  can  be 
determined  very  readily  by  means  of  a  certain  function, 
viz.,  LEGENDRE'S  function,  which  has  been  worked  out  and  is 
tabulated  much  in  the  same  way  as  trigonometric  functions. 
LAPLACE'S  equation 

g-O  (1) 


dx2 

can  be  used  as  has  been  shown  in  exploring  the  space  surrounding 
charged  body. 

With  circular  symmetry  of  the  charged  body  it  is  obviously 
advantageous  to  express  the  equation  in  spherical  coordinates 
(see  Appendix  heading  Partial  Differentiation).  Thus, 

rd2  (rV)       _J_J)/.       dF\  1      d2V 

dr2  sin  6  dd  \Sm      BO/        sin2  6  d<?2 

With  z-axis  as  the  axis  of  circular  sym- 
metry, the  potential  will  be  the  same  for  ,-- 
all  values  of  <£,  as  long  as  r  and  6  are  con- 
stant, as  is  readily  seen  in  Fig.  84. 

Equation  (2)  becomes: 

rd2(rV)    ,       1      < 


(2) 


dr2 


(3) 


This  is  then  an  equation  of  two  inde- 
pendent variables,  r  and  0.  The  general 
method  of  solving  such  equation  is  to  FIG.  84. 

assume  the  solution  to  be: 

V  =  R'6',1  where  Rr  is  a  function  of  r  only,  and  0'  is  a  function 
of  6  only. 

Substituting  in  (3), 


lNoTE. — See  Byerly's  "Fourier's  Series  and  Spherical  Harmonics." 

189 


190  ELECTRICAL  ENGINEERING 

or,  *       .,  d2    .  R'     d  30' 

re'^(rR'}=     -^^sin*-, 

or,  r    a2  (rR'}  I        d  I  .       dO'\ 

W  ~^~      ~  W^e  de  (sm  e  W  <5> 

The  left-hand  term  is  a  function  of  r  only,  the  right-hand  term 
of  6  only. 

In  order  then  that  this  shall  hold  for  all  values  of  r  and  0,  each 
term  must  not  only  be  a  constant,  but  must  be  the  same 
constant. 

Let  this  constant  which  is  entirely  arbitrary,  be  a2, 

-««'-0  (6) 


and  1      d  /sin  6 


Equation  (6)  becomes; 

rS^  +  2r 

dr2  dr 


dr 
The  solution  of  (8)  is  readily  found,  it  is  : 

where 
and 

(9) 


It  is  evident  then  that  rm  and  -^i  are  particular  solutions  of 

equation  (6). 

If  we  choose  for  a2  a  value  which  is: 

a2  =  m(m  +  1), 
then  equation  (9)  is  satisfied,  since 


LEGENDRE'S  FUNCTION 


191 


It  has  been  shown  that  rm  is  a  particular  solution  of  R',  thus 
using  this  solution  at  first,  we  get 

V  =  rmB\ 
Substituting  this  in  equation  (3)  we  get, 


and, 


36 


Equation  (11)  can  be  solved  for  0'. 
We  have, 

d    I  .        a0'\          .        a20' 


Let 


80' 


a:  =  cos  0,  there  sin  0  =  \/l  —  x2'. 

rift' 

In  equation  (12),  is  to  be  determined  --  and  -  ^-' 

ou  do 

80'        86'         8x  86'    .  86' 

^  =  "      smd==  " 


\j  ^/»v  \j  \J  v/**/ 

/  a  a0'\  ax  _  r_a_  /    a0_' 
\ax  a0  /  a0  ~  Lax  \    ax 


a20' 
a02 


(14) 

U^l/ 
r)  f)f 

Substituting  the  value  of  T~T-  from  equation  (13)  and  the  value 

ofj 

of  —j£  from  (14)  in  (12),  we  get: 

a 


Thus  equation  (11)  becomes: 

r)2/?' 
m(m  +  1)  8'  +  ~  (1  -  x 


^-(x  +  x)  =  0, 
+  !)«'  =  0  (16) 


192  ELECTRICAL  ENGINEERING 

This  equation,  which  very  important,  is  called   LEGENDRE'S 
equation. 

It  can  also  be  written: 


™ 


[(1  -  x«)  ~]  +  m  (m  +  1)  tr  -  0  (17) 


since> 


Assume  now  that  0'  can  be  expressed  in  whole  powers  of  x 
multiplied  by  constant  coefficients,  that  is, 

9'  =  2anxn  =  a0  +  aix'  +  a2z2  +  a3z3  +  .    .    .          (18) 
Referring  to  equation  (17), 

~\nf 


(1  -  x2)     -  =  ai  -  aiz2  +  2a2x  -  2azx3  +  3a3xz  -  3a3a:4 


—  (1  -  x2)  —  =  -  2aix  +  2a2  - 

and, 

m  (m  +  1)  6'  =  m  (m  +  I)a0  +  m  (m  +  1)  aiz'  +  m  (m  +  1)  a2^2  + 

m(m  +  1)  •  a3z3  +  .    .    . 

Collecting  the  coefficient  for  similar  powers  of  x  we  get: 
[2a2  +  m(m  +  I)a0]  is  the  constant  term; 
[6a3  —  2ai  +  m(m  +  l)oj  is  the  coefficient  of  x1', 
[  —  6a2  +  12a4  +  m(m  +  1)  a2]  is  the  coefficient  of  x2; 


Since,  from  equation  (17),  each  of  these  coefficients  is  zero,  we 
get: 

m(m  -f-  l)ao  m(m  -\ 


—  2 


***  a  a  i> 

D  0 

m  (m  -f-  1)  a2  +  6a2  _        m  (w  +  1)  ~  6 

It  is  seen  that  if  a0  =  0,  all  the  even  terms  disappear;  if  ai  =0, 
the  odd  terms  disappear. 


LEGENDRE'S  FUNCTION  193 

The  coefficients  are  related  in  a  comparably  simple  manner,  as 
follows: 


I       ON    L'»»\"* 

or,  (fc+l)(fc 

m(m  +  1)  - 

(fc +!)(*- 


From  (20)  it  follows,  that,  if  fc  =  m  -  2, 

(m  -  2+  l)(m  -  2  +  2)  m(m  -  1) 

~  (m  -  m  +  2)(m  +  m  -  2  +1)  am~       2(2m  -  1) 
m(m  -  1)  (m  -  2)  (m  -  3) 


_ 


2.4(2m-  I) (2m  -  3) 
m(m  —  l)(m  — •  2)(m  —  3)(m  —  4)(m  —  5) 

x i _ . L-± £_  ft       .    r»ff> 

2  •  4  •  6  (2m  -  l)(2m  -  3) (2m  -  5) 

It  is  thus  possible  to  express  equation  (18)  as  follows:  6'  =  2 anzn; 
if  the  highest  power  of  x  is  xm,  then  we  get: 


, 


2   (2m  -  1) 
W(m  -  1)  (m  -  2)  (m  -  3)  ] 

2  •  4(2m  -  1)  (2m  -  3) 

where  am  is  entirely  arbitrary,  and  it  is  convenient  to  choose  a 
value, 

(2m  -  1)  (2m  -  3)  (2m  -  5)    ...    1 
am  =  7— 

because,  for  this  value  of  am,  6'  —  1  when  x  =  1. 

.     ,  =  (2m  -  1)  (2m  -  3)  (2m  -  5)  ...  1  r  _  m(m  -  1)    m_2 

m!  2(2m  -  1)  ^ 

m(m-l)(m-2)(m-3)  1 
2-4(2m  -  1)  (2m  -  3) 

Since  6'  is  a  function  of  #,  and  contains  no  higher  power  of  x  than 
>xm,  it  is  customary  to  write,  instead  of  6',  Pm(x),  or  since  x  was 
cos  0,  Pm(cos  6). 

Before  enumerating  some  values  of  6',  recollect  that  (factorial  0)  =  1, 
or  0!  =  1,  or  I?  =  1;  and  since  |1  =  1,    0=1  =  1.     This  is  readily  seen 

In  1 

since  jn_  l  =  n]  forn  =  1,  •=  =  1,     .*.  |0  =  1. 


194 


ELECTRICAL  ENGINEERING 


Example. — Find  P3  (cos  0). 


m  =  3,    .'.P3(z)  = 


(6  -  1)  (6  -  3)  (6  -  5) 
1-2-3 


r  8      3  3-1  ,  -i 
2 '  6  -  lx  . 


Note  that  only  three  terms  can  be  used  in  the  numerator  in 
front  of  the  parenthesis,  -since  the  last  term  must  end  with  1  as 
is  shown  in  equation  (22). 

The  parenthesis  contains  only  two  terms,  because  the  next 
term  would  give  a  negative  exponent,  and  we  have  assumed  that 
the  powers  of  x  are  positive  integer  numbers. 
Thus,  for  m  =  3, 


P3 


Similarly,  for  m  =  2, 
(4  -  1)  (4  -  3) 

1-2 
For  m  =  1, 

Pi(s)  =  —^  x  =  x 


or, 

V  =  A0r°P0  (cos  0)  + 


(23) 


For  m  =  0, 


But  we  assumed  as  a  particular  solution: 

V  =  r"0'  .',  V  =  2AmrmPm(x),  or  2AmrmPm(cos  6)       (24) 


(cos  0)  +  A2r2P2  (cos  0)  + 

A3r3P3  (cos  0  +    .    .    .     (25) 


Referring  now  to  equation  (10),  we  see  that  there  is  also  another 
particular  solution,  namely: 


or, 

_ 


A2P2(cos^) 


(26) 


Before  applying  these  equations  to  some  practical  problems,  it 
may  be  of  interest  to  note  that  the  LEGENDRE'S  function  can  be 


LEGENDRE'S  FUNCTION 


195 


obtained  by  expanding  ^  where  R'  is  the  distance  between  two 
points  (Fig.  85). 


and 


R'  =   \r2  +  ri2  -  2rri  cos  0. 
If  n  >  r,          lf    ;  Ur5        t  __  2r        .T^       A 

where  A  =  (1  +  h2  - 

where  T 


p  =  cos  e. 


FIG.  85. 
Expanding  A  by  the  binomial  theorem,  we  get: 


hp 


Po  + 


-  3p) 

+  /i2P2  +  /i3P3  + 


.    .     (27) 


The  similarity  between  (23)  and  (27)  is  obvious. 

Returning  now  to  the  problem  of  a  circular  wire  carrying 
current,  we  have  shown  that  the  potential  at  a  point  on  the  axis, 
that  is,  r  coincides  with  i/-axis  and  6  =  0,  is  : 


where  r  and  R  are  shown  in  Fig.  86. 
If  R  >  r,  see  Fig.  86,  then 


FIG.  86. 


196  ELECTRICAL  ENGINEERING 


Remembering  that  when  -^  =  K  is  a  fraction 


Since  equation  (23)  holds  for  all  values  of  0,  it  also  holds  when 
6  =  0.  Thus  we  can  readily  determine  the  coefficients  A0,  Ai, 
A  2,  etc.,  which  are: 

Ao  =  2*1, 


A     —    - 

R  ' 

A,  =  0, 
_  .,_,   1 


A4  =  0, 

A6  =  0, 
AT-  + 


y  =  27r7[l  --  ^P!  (cos  0)  +  M  ^3^3  (cos  0)  - 

M  ^5  (cos  0)  +  ^  •  ™P7  (cos  0)  +    .    .    .]       (28) 
If  r  >  R,  then, 


r  ^      !.3B4      l-3-5fl6  -i 

2.7  [1  -  1  +  K  -  -  —  -  +  ^-Q  -  + 

[7?2  P^  z?6  n 

>i^-M^+K6|+    •    •    •]  (29) 


From  equation  (26)  we  get: 

+  ±Pl«*e  +  Al*& 


.*.  Ao  =  0, 

R2 

1  =  2?r/'T' 


LEGENDRE'S  FUNCTION 


197 


A2  =  0, 

A3  =      -  2irl  • 

A,  =  0, 

A  6  = 


/.  V  =  2irl  [^^1  (cos  0)  -  %  ^4PS  (cos  6)  + 


7?6 


-, 

.      ;J     (30) 


As  a  second  application  of  the  use  of  the  LEGENDRE'S  function, 
the  following  problem  will  be  considered. 

Find  the  potential  at  points  outside  of  a  thin  circular  disc,  Fig. 
87,  charged  to  a  certain  potential,  V. 

It  will  be  proven  that  the  distribution  of  the  surface  charge  is  : 


where  Q  is  the  total  charge,  that  is,  the  charge  on  both  sides. 


FIG.  87. 


FIG.  88. 


We  first  calculate  the  potential  at  a  point  PI  on  the  axis 
(Fig. 

/"Vo  =  0  /"*ro  =  0 

2Q 


Q         1r2- 
=  2K  cos     ^T^ 
as  can  be  readily  found  by  simple  integration. 

This  expression  then  must  be  expanded  in  a  power  series. 
This  can  not  be  readily  done,  but  its  derivative  with  respect 
to   r  becomes   a   simple   expression,    which   can   be   expanded, 
the  resulting  series  can  be  readily  integrated.     Thus, 

^.  ["-*?-       -i  r2  -  R21  =  Q 

drl2R  C         r2  +  R21  R2  +  r2 

If  R  >  r,  then 

Q  ^H        £,J1       J6,  1 

fia  +  r2  B2L        R2~*~  R4       R&^  '    '    '   J 


198  ELECTRICAL  ENGINEERING 

Integrating, 


R        R 


-&-•  •  -+c} 

OH  J 


For  r  =  0,  i.e.,  on  the  disc,  and  the  potential  of  the  disc  will  be 
proven,  to  be  ^  '  D"' 

•  r  -  - 
"  2 

•  7  -if--  -  -4-  —  4-  —  1 

"  R  12       R  +  3^3  +  5^5  " 
when  r  >  R,  it  is  found  in  a  similar  way  that: 

'-§[?-£+£-£+•••] 

Equation  (31)  is  similar  to: 

7  =  A0r0P0(cos  0)  +  Air!Pi(cos  0)  +  A2r2P2  (cos  0)+  .    .    . 

''•  Ao  =      ''  Al  =  ~''A2  =  °>As  =      '^)A^  °'  etc' 


•*•  V  =  |[IP°(COS  6)  "  iPl(cos  ^)  +  3^^2(cos  0) 


Equation  (32)  is  similar  to: 

_  A0P0cos0     '  AiPi  (cos  0)       A2P2  (cos  0) 

fl  f2  r3  -     '      •      • 

/.  A0  =  5-  B,  A!  =  0,  A2  =      -  ^  y,  A3  =  0,  A4  =  ^  •  y,e 


CHAPTER  XVII 
DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID 

If  an  ellipsoidal  thin  shell  is  formed  by  two  similar,  similarly 
situated  ellipsoids,  and  the  charge  per  unit  volume,  p,  is  constant 
in  the  shell,  then  the  force  at  any  point 
inside  the  ellipsoid  is  zero,  that  is  the  poten- 
tial is  constant.  The  outer  surface  is  an 
equipotential  surface.1 

To  prove  this,  consider  the  attraction  at 
o  of  the  two  masses  at  A  and  B,  Fig.  89.  FIG.  89. 

The  volume  at  A  is  rz  du  dr         .'.  charge,  q  =  pr2  dudr. 
The  volume  of  B  is  n2  du  dr       .'.  charge  q'  =  pr^  dudri 

.'.  The  attraction  of  A  at  0  is  -^  =  pdu  dr. 

The  attraction  of  B  at  0  is  —2  =  pdu  dri. 

But  from  geometry  it  is  known  that  with  two  ellipsoids,  one  of 
axes  a,  b  and  c,  and  the  other  of  a  (1  +  a),  b(l  -f  a)  and  c(l-f  «), 
that  is,  with  two  similar,  similarly  situated  concentric  ellipsoids, 
dr  must  always  be  equal  to  dri.  Thus  the  attraction  at  0  must 
be  zero. 

In  the  case  of  a  conducting  ellipsoid  charged  with  electricity, 
the  charge  is  confined  to  the  surface  and  the  distribution  will  be 
shown  to  be  such  as  is  represented  by  the  thickness  of  the  shell 
in  Fig.  89.  It  is  greatest  where  the  curvature  is  greatest  and 
least  on  the  flat  point  of  the  surface. 

The  problem  then  is  to  express  the  thickness  of  the  shell  in 
terms  of  a  variable  surface  charge,  cr. 

The  volume  of  the  shell  is  evidently  =  %irabc  [(I  +  a)3  —  1]; 
considering  uniform  volume  charge,  the  total  charge  is: 


1  NOTE.  —  See  "Analytical  Statics,"  vol.  II,  by  ROUTH. 

199 


200  ELECTRICAL  ENGINEERING 

But  a  =  pd,  where  5  is  the  variable  thickness  of  the  shell, 

:.Q- 


or 


3QS 


J  s 
QS 


FIG.  90. 


But  the  thickness  of  the  shell  5  can  be  ex- 
pressed as  the  distance  between  two  parallel 
planes  going  through  any  point  of  the  shell. 
We  have  from  geometry  (see  Fig.  90)  that 
the  distance  from  the  center  of  an  ellipsoid  to 
a  tangent  plane  is  : 

P  =  ~  (1) 


/£       g»       *» 

\  a4  "•"  fe4  "*"  c4 


Neglecting    infinitesimals    of    higher    order    than    the    first, 
d  =  p(l  +  a)  -  p  =  pa. 

Qp 


.  .  a  = 


4irabc 


y  +  a  +  l) 


;  or  at  the  limit  «  =  0, 


(T    = 


4:irabc 

Consider  now  a  very  thin  flat  elliptic  disc  in  the  x  —  y  plane 
(c  is  small)  we  have  from  (1) 


Q 


o 


when  c  approaches  zero, 


47Ttt& 


/         x2       »• 

V J  -  *  -  v 


As  a  consequence  for  a  circular  disc, 

Q 


a  = 


-  r 


DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID     201 

where  R  is  the  radius  of  the  disc  and  r  the  particu-  P 

lar  distance  from  the  center,  where  a  is  the  surface 
density  on  the  disc. 

To  find  the  potential  of  the  circular  disc,  we  calcu- 
late the  potential  at  a  point  on  the  axis,  Fig.  91.  FIG.  91. 

rdrQ 


A  =  °    2Trrdr2<r  f° 

V     =     -  —r--  -  ~     =     —    4-7T 

Jr-R      Vx*  +  r*  JR 


4irR\/R2  - 


r* 


=   __Q   T°  __          rdr 
'  RJR  V(R2-rz)(x 


In  this  equation,  x  is,  of  course,  a  constant,  being  the  distance 
from  the  disc  at  which  the  potential  is  to  be  determined : 
On  the  disc,  x  =  0, 

Q    C°        rdr 


Q        dr 


0  Q 


Incidentally,  since  the  capacity  is  ^,  it  follows  that  the  capacity 

2 

of  a  disc  is  -  R,  which  is  2/Tr  times  that  of  a  sphere  of  the  same 

7T 

radius. 


FIG.  92. 

Potentials,  Outside  and  Inside,  and  in  the  Body  of  a  Spherical 
Shell. — Let  the  uniform  charge  per  unit  volume  of  the  mass  of 
the  shell  be  p,  and  the  inner  radius  r0  and  the  outer  radius  R, 
Fig.  92. 

The  area  of  the  shaded  surface,  Fig.  92,  is  r^<p  •  rA0 
=  r  sin  6  •  A^>  •  rA0; 


202  ELECTRICAL  ENGINEERING 

the  volume  of  an  element  of  thickness  Ar  is : 

r2  sin  0A0A0Ar. 

If  p  is  the  charge  per  unit  volume,  then  the  charge  on  the 
small  volume  is: 

q  =  pr2  sin  0A<£A0Ar. 

Thus  the  potential  function  at  P  due  to  the  charge  on  the  small 
volume  is: 


V  =  ^  but  a  =  Vri2+  (c  -  r  cos  0)2 

=  Vc2  -f  r2  sin2  0  +  r2  cos2  0  -  2cr  cos  0 


=  V  c2  +  r2  -  2  cr  cos  0  ; 

or,  a2  =  c2  +  r2  -  2cr  cos  0  (1) 

pr'sinftfrdfrfr 


a         r  =  ro      ^=0        ,  =  0         c2  +  r2  =  2cr  cos  0 
From  (1),  a2  =  c2  +  r2  -  2cr  cos  0, 

.*.  2ada  =  2cr  sin  0d0 

sin  AM  =  ^  (3) 

cr 


Substitute  (3)  and  (1)  in  (2), 
r=R 


_     fr 

Jr 


pr'adadrd* 


"     prdadrd* 


^2"  f 

=  Q      J 
fr  =  R    r?  =  2* 

r[(c  +  r)  -  (c  -  r)]  drdS  (4) 

C  Jr  =  ro    Jv  =  Q 
r  =  R    r 

I 

=  ro    Jv 


^  I         2irr*dr 


=  47T5  /^!3  -  r03\  =  p(volume  of  shell)  =  Q 
c    \        3       /  =  c  =   c 


(5) 


If  point  P  had  been  inside  of  the  shell,  then  the  limits  of  inte- 
gration of  a  would  be  r  —  c  and  r  +  c. 


DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID    203 
.'.  Equation  (4)  would  be: 

X"»       Tp  X"»         _.   O 

F  =  2   (          |         r[(r  +  c)  -  (r  -  c)]drd<p. 

C   Jr  =  ro     Jv  =  Q 

2     rr=R   r<p=2* 

I      crdrdp 


Jr- 


-B 


(8) 


which  is  independent  upon  c,  the  position  of  the  point  P. 
Thus  the  potential  is  constant  inside  of  a  hollow  sphere. 


FIG.  93. 


If  the  point  had  been  in  the  body  of  the  shell,  Fig.  93,  then  the 
potential  would  be  the  sum  of  the  potentials  due  to  the  mass 
outside  and  inside  of  the  spherical  surface  which  contains  P. 


The  field  intensity  or  potential  gradient  is 

dv 


(The  signs  should  all  be  reversed  for  gravitational  potentials.) 
In  the  case  of  the  point  being  outside  the  sphere, 
dV      Q        4™ 


dc 


3c2 


-  r03) 


and 


2Q       STTP  (R*  -  r03) 
c3  :  3c3 


In  the  case  where  the  point  is  inside,  it  is: 


(8) 
(9) 


204  ELECTRICAL  ENGINEERING 

where  the  point  is  in  the  shell  then  : 

^Z  o       f       ?cf       W-l 

"  dc  =          ^p  L"     3    "   3cJ 

rV  _     1  _ 
c3 


2r02 


(10) 


dc 


31 


c3 


Problem.  —  Plot   the   potential,    the   potential   gradient,    and 
d2V 


•j-z 
dc2 


when  V  —  1  at  the  center; 


ri  =  1; 

r-o  =  0.5 

in  the  case  shown  in  Fig.  94. 

For  a  full  discussion  see  WEBSTER'S  "Electricity  and  Magnetism." 


FIG.  94. 


FIG.  95. 


Potential  Outside  of  a  Non-conducting  Charged  Oblate  Ellipsoid, 
—  Let  the  equation  of  the  oblate  ellipsiod,  Fig.  95,  be: 


x2       y2      z2 

~o   ~I          9   ~T~    ~o    = 

o2       a2       c2 


DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID    205 


Let  the  total  charge  of  the  ellipsoid  be  Q,  and  the  potential  on 
the  surface  be  VQ. 

The  surface  intensity  at  the  element  ring,  generated  by  ds, 
Fig.  96,  revolved  about  the  z-axis,  has  been  proven  to  be: 

pQ 

(T     - 


— X 


FIG.  96. 
where  p  is  the  distance  from  the  origin  to  ds,  and 


5  +  *  +    ~       + 


where 


Q 


,2  ~2 


From  (1), 

a2  ^  c2  ~ 

Differentiating,          2r0  drQ       2zdz  _ 

~2  I  ^2         ~~    U> 


or, 


(2) 


(3) 


206  ELECTRICAL  ENGINEERING 

The  potential  at  P  on  the  axis  due  to  the  ring-shaped  element 
surface  is: 

(4) 


(r  - 

Substituting  (2)  and   (3)   in  (4)   we  get:  the  potential  at  P 
due  to  the  whole  ellipsoid 

Qdz 


•r, 


2cV(r-z)2 
From  the  equation  of  the  ellipsoid, 


substituting  in  (5), 

"•=£ 


(5) 


£ 


f_c  2V  -  (a2  -  c2)  z2  -  2rc2z  +  c2  (a2  +  r2) 

,   (a2  -  c2)z  +  re2 1 c 

, sin"1  - — .  =         = 

2V a2  —  c2  caVa2  -  c2  +  r2J  _c 

Q          f.     .     a2-c2  +  rc  .     1-a2  +  c2  +  rc]/ 

.  sm"1 x  =  —  sin"1  — .  — :    (6) 

2V  a2  -  c2[          aVa2  -  c2  +  r2  a-ya2  -  c2  +  r2J 

To  find  the  potential  at  a  j>oint,  like  PI,  which  is  not  on 
the  z-axis,  LEGENDRE'S  function  may  be  employed,  and  the 
equation  (6)  is  to  be  expanded  into  a  series  in  the  terms  of  r. 
In  order  to  obtain  an  expression  which  may  be  easily  expanded, 
differentiate  (6)  with  respect  to  r,  expand  the  result  into  a  series, 
and  then  integrate  the  series.  Thus  differentiating  (6), 


dVp  Q 


(7) 


dr  (a2  -  c2  +  r2) 
Expanding  (7), 

dVP          -Q  r  r2  r4  _r^_  l 

dr        a2-c2L         a2-c2~i  (a2  -  c2)2    "  (a2  -  c2)4  ±    '        'J1 

when  c  <  r  <  \/a2  -  c2  (8) 

V  Q     -[_  c  ,    y(      _c         V  _ 

"         2        2  2        2  \Va2  -  CV 


DISTRIBUTION  OF  CHARGE  ON  AN  ELLIPSOID     207 
For  a  point  on  the  surface,  i.e.,  when  r  =  c, 


.-.  C  =  V~      +  tan-'  -7=  (10) 

^  V  a2  —  c2 

Since  FPl  is  a  function  of  VP  and  /£,  the  solution  for  VPl  takes 
the  following  form: 

VPl  =  Ao  +  AiriPi(cos  5)  +  A2ri2P2(cos  0)  + 

A3ri3P3(cos  0)  +  .... 
When 
0  =  0,  n  =  r,  Pi  =  P2  =  P8  =   -    -    .   «  1,  and  FPl  =  FP. 

^°  +  -  -  tan-1  -  7J= 

Q        QVa2  -  c2  \/a2  -  c2J 


a2  -  c2! 
=  0; 
.  U-- «- 


(a2  -  cV 

1  c        A      P^cosfl) 

tan   L  — 7=  =  I 7—^- ^r 

(a2  —  c2) 


P3(cos  e)      ,         P5(cos  0)       ,  ,     P7(cos  6)      7  _ 

I  v /     ~*  3 * L-    M  5  I      x  y     ,«   7    _i_  I  , 

I         O  /       9  9\  9*1  F*  /       *>  9\  *?     '    1  I^     T  /       9  9\  A     '   4  I  •          •          •  I 

3  (a2  —  c2)2  5  (a2  —  c2)3  7(a2  —  c2)4 

which  is  applicable,  when 
When 


expanding  (7), 


Whenr  =   c»,  VP  =  0,  .'.  (7  =  0. 
And 


_      Q    ,       ii)    ,   A2P2(cos0)   ,      33  , 

I  ™  n"       "  r!2  n3  4         -  •  •  •  • 


208  ELECTRICAL  ENGINEERING 

When  0  =  0,  PI  =  P2  =  P3  =  .    .    .   =  1,  rl  =  r,  and  VPl  =  VP) 
/.  ^o  =  Q; 
A!  =  0; 
4  Q(a«-c»). 

"T""; 

A*  =  0; 


=  0; 


a2  -  c2)  Ps  (cos  ^)    .    (a2  -  c2)2  P4  (cos  0) 
~~  ~ 


(a2  -  c2)3  P6  (cos  6>) 

7ri7  - 

which  is  applicable,  when 

ri  >Va2-c2.  (12) 

(Two  similar  series  can  be  derived  for  an  oblong  ellipsoid. 
For  this  and  the  potential  at  a  point  inside  an  ellipsoid,  see 
W.  E.  BYERLY'S  "Series.") 


CHAPTER  XVIII 
CONCENTRIC  SPHERES 

Fig.  97  represents  a  system  of  concentric  spherical  shells.  It 
is  desired  to  find  the  potential  at  any  point  in  the  medium  (which 
is  assumed  free  from  charge). 

Since  we  are  dealing  with  spherical  bodies  and  since  the  body 
is  symmetrical,  indeed  a  sphere,  LAPLACE'S  equation  in  spherical 
coordinates  becomes : 

d*V 

aPPendlx)  (1) 


FIG.  97. 

To  solve  this  equation,  one  first  ascertains  if  the  relation 

dV  =  A 
dr  ~  r2 

is  satisfactory.     (We  may  well  assume  this  solution,  since  it  can 
be  expected  that  the  intensity  or  force  on  unit  charge  varies 
inversely  as  the  square  of  the  distance.) 
Then, 

d^V        _2A 

dr2  ~~          r3' 

Substitute  in  (1)  to  see  if  the  solution  satisfies  the  equation 

-~  +  l£  =  o,Q.E.D. 

Thus,  37  _  A 

~dr  ~  7* 
209 


210  ELECTRICAL  ENGINEERING 

satisfies  the  equation  (1). 

or,V=-      +  B  (2) 


Or  we  might  have  solved  the  equation  as  follows : 

Let  _  dV       .  dW  _  dy 

y  ~~   dr'       '  dr2       dr 

.-.*+*.-« 

-  J*-dr  -log  r2  A  A 

.  .  v  =  Ae  =  Ae  =  — , 1)  —  -$• 

e  log  r*       r2" 


Or  again  we  might  have  developed  the  equation  directly,  without 
using  LAPLACE'S  equation,  by  assuming  a  positive  charge  Q  on 
the  inside  sphere. 

The  intensity  of  the  field  at  a  point  in  the  medium  at  a  distance 
r  is  then  by  GAUSS'S  theorem: 

A~n      n 

R  = 


.'.  V  =  -   \  Rdr  =   +  ^  +  B  (3) 

an  equation  of  the  same  form  as  (2). 

Referring  to  equation  (2),  let  Vi  be  the  potential  of  the  inner 
sphere  of  radius  r\  and  Vz  that  of  the  outer,  then, 


and'  F2  =  -  ™  +  B. 


r2       r  rir2 


r2  — 


r2  -  ri        r 
where  ri  <  r  <  r2 


CONCENTRIC  SPHERES  211 

To  determine  the  meaning  of  B  assume  that  the  outer  shell  is 
grounded,  or  which  is  the  same,  at  zero  potential,  then 


and 

from  (4), 


.   v  tfi          D    .    K  rl 

.   .    V  2    =    0    ~    ~~  -    -T  jD.    .   .  £>    =     —    -  -     V  i. 

r2  -  ri  r2  -  n 

From  (5),          y  =       Vi      n  r2  _    _TI  __ 

r2  -  ri     r          r2  -  ri     * 


=  _FVi_  rra  _11==  JV 
r2  —  ri  Lr         J      r2  — 


-  r 


7*1         r 

If  the  outside  sphere  is  very  far   off   so   that   r2   approaches 
infinity  and  F2  zero,  then, 

Vz  =  0,  r2  =   oo  ; 

72  =  o  =  -  —  +  5,  /.  5  =  0. 


r2       ?  r  r 

The  potential  gradient  in  the  space  between  the  conductors  is: 

7.-F. 


It  is  the  greatest  at   the  surface  of  the  inner  sphere,  where 
r  =  ri. 


r2  — 


The  potential  gradient  at  the  inner  surface  of  the  outer  conductor 
is  evidently  : 


r2  —  ri       r2 
Referring  to  equation  (7), 


R  =  4^ri  =  ^J  equating  to  (8), 
Qi    _  7i-72  ra. 


r2  - 


14 


212  ELECTRICAL  ENGINEERING 

Example. — Calculate  the  average  potential  gradient  in  the 
space  between  two  concentric  spheres  separated  by  a  distance  of 
2  cm. 

Assume  that  the  potential  gradient  at  the  surface  of  the  inside 
conductor  is  100  electro-static  units  per  centimeter,  that  is,  just 
about  on  the  point  of  glowing. 

Consider  a  concentric  sphere,  Fig.  98,  the  inner  sphere  of  which 
has  a  charge  Q\  and  the  outer  a  charge  Qo  =  $2  +  Qs- 


FIG.  98. 

Evidently,  Qo  =  Q2  +  Q8. 

Since  all  tubes  of  force  beginning  at  the  surface  of  the  inner 
conductor  terminate  at  the  inner  surface  of  the  outer  conductor, 
it  is  evident  that  the  charge  Qz  =  —  Qi- 

.'.  Qo  =  -  Qi  +  Q8. 

The  potential  at  a  point  outside  of  the  outer  conductor  is 
thus,  from  (6), 

T7     Qo     Qs-Qi     , 

y  =  ^_  =  ^         *_,  where  r  =  r3. 
r  T 

Since  the  capacity  of  an  electric  field  is  the  ratio  between  the 
charge  on  the  positive  boundary  and  the  potential  difference 
between  the  boundaries, 

c  - 


ri-F2 
Thus  horn  (9), 

c==Ii^Z_2.rir2.        1 

r2  —  ri  Fi  —  K  2       /  2  —  / 1 

The  capacity  of  the  inside  sphere  alone  is  ri. 

Capacity  of  concentric  spheres  _    r^ 

Capacity  of  inner  sphere  r2  — 


CONCENTRIC  SPHERES  213 

If  the  thickness  of  the  dielectric  is  small  compared  with  the 
radius,  then: 

C  =  r,  ^±^  =  T4'  where  S  =  r,  -  r,. 
6  d 

4?rri2   _  area  of  sphere 


as  a  limiting  case,  where  TI  =  r^  =  °°  we  get  parallel  plates,  and, 

area  on  one  plate 


47r(distance  between  them) 

The  capacity  is  expressed  in  cm.  not  in  farads.     To  get  the 
capacity  in  farads  divide  C  by  9  X  10.  u 
The  energy  input  to  a  condenser  is: 

W  = 


Thus  the  energy  stored  in  the  field  between  two  concentric 
spheres,  is: 


Infinite  Parallel  Planes.  —  LAPLACE'S  equation  applies  in  this 
case  so  long  as  there  are  no  charges  between  the  condenser  plates, 


_  d*V 

dx2  "  dy2  ~~  dz2  = 

Since  the  field  depends  upon  the  distance  between  the  plates  only, 
that  is,  upon  one  of  the  coordinates  only,  we  get, 

w  _      .  dv  __ 

"T     "^    —    U,    .   .      j        —    L/o 

dx2  dx 

and  V  =  C0x  +  d.  , 

If  the  charge  on  plate  A  (Fig.  99)      A      ifc^T       |          Ql 
is  Qi  and  the  potential  FI;  and  the  , 

charge  on  plate  B  is  $2  and  the  poten-      B  —  -  -  1  -  Q* 
tial  Vz',  and  if  the  distance  between 
the  plates  is  d] 
We  have: 

Vi  =  0  +  Cj, 
and 

F2  =  C0d  +  Ci. 

Subtracting, 

Vi  -  Vz  =  -  Cod,  or,  C  =  -  (Fl  "T  V^. 


214  ELECTRICAL  ENGINEERING 

:.V  —-^  -x  +  Ci;  or,  since  V1  =  d, 


d 
The  potential  gradient,  that  is  the  potential  drop  per  cm.  is: 

«=-£=^ 

It  is  constant  all  through  the  dielectric. 

The  total  outward  flux  from  A  is  47rQi,  one-half  of  this  enters 
the  space  between  the  plates.  The  inward  flux  to  B  is  ^irQi, 
and  one-half  of  this  is  added  to  the  flux  from  A.  Thus  the  total 
flux  in  the  space  between  the  plates  is: 


But  the  charge  on  A,  Qi,  must  be  numerically  the  same  as  that 
on  B,  Q%,  since  all  tubes  of  force  leaving  A  enter  B,  thus  Qi  =  $2, 
numerically,  but  of  course  of  opposite  sign,  which,  however,  is 
taken  care  of  in  the  above  discussion. 

Thus  the  total  flux  in  the  gap  is  4?rQ,  where  Q  is  the  charge  on 
one  of  the  plates. 

47T0 
.'.  R,  the  intensity  of  the  field,  is  — j-»  where  A  is  the  area  of  one 

side  of  the  plate. 

And'  G  =  R  =  ^; 

or  from  (1), 

4rQ       Fi- F2     .   r  _         Q  A. 

A  d      "  '  '  L     =  Fi  -  72  " 


This  could  have  been  calculated  in  still  another  way. 
Since  D 

V  =  -  fRdx  =  -  X 

For  x  =  0,  V  =  7i;  /.  Ci  =  7i. 


for  ,  ,,       T7      .   T7        T7 

oj  =  d,  V  =  72;  .  .  Vz  =  Vi  --  T—  d, 


CONCENTRIC  SPHERES  215 


1s  .  2  .. 

If  the  plates  are  separated  by  uniform  insulation  of  specific 

.,       v    ,,  .,     .    KA  KA 

inductive  capacity,  A,  the  capacity  is  -r—,  cm.,  or  .    ,  Q   1()ll 

farads. 

If  the  dielectric  consists  of  several  layers  of  different  specific 
inductive  capacities  then  one  can  consider  that  the  condenser  is 
made  up  of  a  number  of  condensers  in  series  and  the  capacity 
of  each  is: 

KiA 

Ci  =  -7—r>  etc. 
47rdi 

The  total  capacity  is  obtained  from  the  well-known  relation: 


1     1 

C  ~  Ci  ' 

1 

f  7T  +     •    •    .    ,  or, 
C2 

1 

1 

,  1  , 

±irdi        4ird2 

Ci 

F  c2  +  •  • 

V       *      V       *      '      •      • 
AI         A2 

All  these  formulae  are  approximate,  however,  since  no  allow- 
ance has  been  made  for  the  effect  of  the  edges,  but  the  plates 
were  assumed  to  be  infinite. 

Concentric  Cylinders. — LAPLACE'S  equation  can  again  be  used 
if  it  is  assumed  that  there  are  no  charges  between  the  cylinders. 
Moreover  since  we  are  dealing  with  cylinders,  it  is  best  to  put 
LAPLACE'S  equation  in  cylindrical  coordinates.  Thus  we  have: 

~dr*+r   ~fo   ~~ 
let  y  =  -r->  then  (1)  becomes  -j-  H —  y  =  0. 

The  solution  of  this  equation  is 

-f*L  =     A      =A 

y  ^^  -.  log  r     "     M 

I 

dV      A 


216 


ELECTRICAL  ENGINEERING 


To  determine  the  integration  constants, 

let         V  =  Fi,     r  =  r,  (Fig.  100) 
and       V  =  F2,  r  =  r2. 
Then,  Vi  =  A  log  (n)  +  B, 
and      Tr2  =  A  log  (r2)  +  5. 

.'.  Fi  -  F2  =  A  (log  ri  -  log  7-2)  =  A  log 


and, 


><*© 


FIG.  100. 

The  potential  gradient  or  the  intensity  of  the  electrostatic 
field  is: 

dV  V,   -    VZ    /1\  47TW  2^ 


where  Qi  =  charge  per  unit  length  of  conductor,  and  Z  =  length 
of  conductor. 


per  centimeter  length  of  conductor. 


The  potential  gradient  is  the  greatest  at  the  surface  of  the  inner 
conductor,  where  it  is: 

1     7,  -  7i 


Graded  insulation  between  the  conductors. 

In  order  that  G  may  be  constant  at  all  points  of  the  dielectric 
it  is  evident  that  the  specific  inductive  capacity  must  be  the  high- 
est at  the  inner  conductor,  and  be  inversely  proportional  to  the 
distance  from  the  inner  conductor. 


CONCENTRIC  SPHERES  217 

Let  the  specific  inductive  capacity  be  expressed  by  the  follow- 
ing formula: 

K  =  ~,  where  a  is  a  constant. 

With  a  charge  Q  on  the  inner  conductor,  the  flux  per  centimeter 
length  is  4irQ,  thus  the  force  on  unit  charge  is: 


_  2Q 
K2irr  ~  Kr 


=-    C^dr=-    (^dr=-    f* 
J  Kr  J    ar  J    a 


dV       2Q 

G  =  ——  =  --  =  constant. 
dr          a 

The  same  result  could  have  been  obtained  directly  from  (2), 
which,  in  the  general  case  when  Kl,  becomes: 

R        4,0        2Q 


K2irr       Kr 


a 

K  =  -> 
r 


Substituting 

R  = 

G) 


20 

R  =  -      -  =  constant,  Q.E.D. 


CHAPTER  XIX 
CYLINDRICAL  CONDUCTORS 

Line  Charge.  —  Assume  that  the  conductor  which  is  perpendicu- 
lar to  the  page  is  infinitely  long  and  its  diameter  so  small  that 
it  may  be  considered  as  line,  and  let  the  charge  per  unit  length 
beQ. 

The  electric  field  is  then  represented  by  radial  lines  in  planes 
parallel  to  the  page  or,  which  is  the  same,  at  right  angles  to  the 
axis  of  the  conductor. 

The  intensity  of  the  field  at  a  point  P,  Fig.  101,  is  obviously: 

2Q 


And  the  difference  in  potential  between  two  points  PI  and  P  is: 

-    C^  dr 
Jhi    r 

-  2Q  [log  n  -  log  fcj  =  2Q  log  (1) 


FIG.  101. 

Two  equal  but  opposite  line  charges  separated  by  a  distance  2hi: 
Let  A  and  B  (Fig.  102)  be  the  locations  of  the  line  charges. 
The  difference  in  potential  between  0 — midways  between  the 
charges — and  P,  due  to  the  charge  on  A  alone,  is  and  has  been 
shown : 

Vp  -  V.  =  2Q  log  £  (2) 

The  difference  of  potential  between  o  and  P  due  to  the  line  charge 
—  Q  on  B  is  obviously, 

Vp  -  V,  =  -  2Q  log  J-1  (3) 

TZ 

218 


CYLINDRICAL  CONDUCTORS 


219 


Thus  the  difference  of  potential  between  0  and  P  due  to  both 
line  charges  is: 

V9  -  V0  =  (2Q  log       -  log     )  =  2Q  log     .  (4) 


Referring  to  equation  (2)  or  (3),  if  P  lies  midway  between  A 
and  B,  so  that  r\  =  r2  =  hi,  then: 

Vp  -  V0  =  0, 

thus  as  long  as  the  charges  are  equal  and  opposite,  the  potential 
at  0  is  zero,  which  would,  of  course,  have  been  concluded  without 
proof. 

V  =  2Q  log  ^  (5) 

where  V  is  the  potential  of  P  due  to  the  charges  on  both  lines. 
From  (5),  follows 


T2  -- 

—  =  €2Q  =  a  =  a  constant 


(6) 


for  all  surfaces  of  potential  V. 

Equation  (6)  represents  a  circle,  defined  by  the  following  relation : 

~OA  X  OB  =  R2  (7) 

referring  to  Fig.  103,  where  0  is  the  center  of  the  circle,  A  and 
B  Fig.  103)  are  called  the  inverse  points,  and  Of  the  center  of 
inversion. 


FIG.  103.  FIG.  104. 

To  prove  that  equation  (6)  represents  a  circle  refer  to  Fig.  104. 


4- 


or. 


O, 


220  ELECTRICAL  ENGINEERING 

which  is  the  familiar  equation  of  a  circle  having  a  radius  of 

o  __  , 

"1-C2 

and  its  center  at  a  point  whose  coordinates  are: 


"1-C2        from  A; 
=  0, 


/.  OA  X  OB  = 


from  B. 


=  R2; 


(1  -  C2)2 
thus,  equations  (6)  and  (7)  are  proved. 

The  ratio,  —  >  can  be  expressed  by  a  simple  equation  involving 

h,  the  distance  of  the  center  from  the  neutral  plane,  and  the 
radius,  R. 


FIG.  105. 
Referring  to  Fig.  105. 

R2  =  OA  X  OB  =  (h  -  hi)(h  + 
hi  = 


or,  i  =         2  -  R2 

But  triangles  OPB  and  OP  A  are  similar,  since 
OP2  =  OA  X  OB; 
-  _T?_       Tl  , 

'OP  ~  OA' 

r,  =  OP  =       R       =  B 

ri  ~  OA  ~  h  —  hi  ~  a 


(8) 


(9) 


CYLINDRICAL  CONDUCTORS 


221 


Substituting  (8)  in  (9), 

rs  = R =  R(h  + 

ri  ~ 


h- 

We  can  then  determine  the  potential  of  a  circle,  or,  which  is 
equivalent  in  this  case,  a  cylinder,  whose  center  is  h  cm.  from 
the  neutral  plane  and  whose  radius  is  R,  as 


Similarly  the  potential  at  a  circle  around  the  negative  charge 


72  =  -  2Q  log 


R 


(12a) 


/.  V  =  Vl  -  72, 

that  is,  the  potential  difference  between  the  two  cylinders  is: 

h  + 


4Qlog 


R 


(13a) 


For  the  sake  of  convenience,  will  be  added  other  expressions  for 
Vi,  Vz  and  V,  involving  hi,  and  R  instead  of  h  and  R. 

From  (8),  h2  =  R2  +  /U2, 

which,  substituted,  gives 

Vl  =  2Q  log  — 


R 


-  -2Qlog- 


R 


and, 


=  4Qlog- 


4-  V  hi2  + 


(126) 
(136) 


It  is  now  evident  how  we  can  go  from  line  charges  to  charges 
on  actual  conductors.  It  has  been  proven  that  the  equipotential 
surfaces  around  the  line  charges  are  cylinders  and  hence  if  circular 
cylinders  be  substituted  for  the  circles,  the  distribution  of  the 
field  is  not  affected. 

The  capacity  per  centimeter  length  of  two  such  metal  cylinders 
(that  is,  of  the  double  conductor)  is  : 


4  log 


h  + 


=  cm. 


(14) 


R 


222  ELECTRICAL  ENGINEERING 

CW---  -f-    -/===  farads  (15) 

9  X  10"  4  log  *     ^~ 

or,  Cm_/.  per  1000  ft.,  of  circuit  (double  conductor) 

m-f.  (16) 


R 

'where  logic  means  the  ordinary  logarithm — not  the  natural 
logarithm — h  is  half  the  distance  between  conductors,  and  K 
the  specific  inductive  capacity. 

If  E  is  the  effective  value  of  the  alternating-current  line 
voltage,  then  the  charging  current  per  1000  ft.  of  double  con- 
ductor is  readily  proven  to  be: 

Cm.f. 

The  capacity  to  neutral  is  obtained  directly  from  (lla)  and  is: 

c=  1 


n,      h  +  Vh2  -  R2 
2  log  -       — gr- 

It  is  thus  seen  that  the  capacity  to  neutral  is  twice  as  great 
as  that  between  the  lines. 

This  results,  of  course,  in  the  same  charging  current  as  in  the 

E 

first  case,  since  in  this  case  the  voltage  is  -^-     Thus  the  capacity 

of  1000  ft.  of  one  wire  to  neutral  or  ground  is: 

Cm-f.  =  ~     —j--    — ,  =  m-f.   per   1000  ft.  of  transmission. 

logio  -     ~/j>~ 

Two  Parallel  Cylindrical  Conductors  of  Different  Diameters 
but  Equal  and  Opposite  Charges. — Since  OA  X  OB  =  Ri2  and 
(FB  X  WA  =  R22,  we  have 

a(a  +  2hi)  =  Ri2,  or  a  —  —  hi  +  \/h\2  +  R\2  (1) 

and          -'-   '   2hi)  =  Rz2,  or  0  =  -hi  +  V/^F^T2  (2) 


and      ^  F,=  -2Qlog^=  -2Qlog|2 


(3) 


CYLINDRICAL  CONDUCTORS 


223 


Substituting  (1)  and  (2)  in  (3), 
Vi  -  V2  =  2Q  log 
and 


C  = 


Q 


(-  hi '+  Vfti2  +  #i2)  (-  fti  +  \//ii 
1 


2  log 


RiR: 


-  hi  +  Vfti2  +  Ri2)  ( -  fti  +  Vfti2  +  #22) 
To  obtain  an  expression  in  terms  of  h  and  R,  instead  of  hi  and 
R,  from  Fig.  106  we  have: 

(3  =  2h  -  2hi  -  a  (4) 

.'.  /?  +  2hi  =  2h  -  a  (5) 


FIG.  106. 

Substituting  (4)  and  (5)  in  (2), 

(2h  - 

Solving  (1)  for  2hi  and  substituting  it  in  (6), 

a 


(6) 


or, 
or, 


(2h  = 


2ha2  + 


—  a)(2h  —  a)   —  R%  , 
« 

-  #i2  -  4/i2)  a  +  2hRS  =  0, 
-  V(Ri2  -  R*2  +  4/i)2  -  16ft2 


4ft 


where   the   sign   in   front   of    the   radical   is   minus  —  not   plus, 
because  a  =  0  when  Ri  =  0.     Similarly, 


4ft 


.'.  C  = 


2  log 


_  __ 
(RiR*\ 


224 


ELECTRICAL  ENGINEERING 


2  log 


4/i2-  (Ri2  +  R22)  -  V  16/i 4  - 
which  becomes: 

C  =  - 


4  log 


h  +  \A2  -  R2 
R 


if  R  is  substituted  for  both  RI  and  Rz,  a  result  obtained  before. 
Construction  of  Equipotential  Surfaces  around  a  Cylindrical 

Conductor,  Charged  to  a  Certain 
Potential,  V. — Let  the  distance  be- 
tween the  center  of  the  conductor, 
Fig.  107,  and  ground  be  h,  and  the 
distance  of  the  equivalent  line  charge 
above  ground  be  hi. 

Since  the  ground  is  an  equipotential 
surface,  it  is  evident  that  the  problem 
will  in  no  way  be  affected,  if  a 
second  conductor  with  a  charge  —  Q 
be  placed  equidistant  below  the 
ground  surface,  and  the  equipotential 
surfaces  around  A  be  considered  as 
due  to  a  positive  charge,  Q  at  A,  and 
an  equal  but  opposite  ("image") 
charge  —  Q,  at  the  inverse  point  A'. 
Suppose  that  it  is  desired  to  draw  the  equipotential  surface 
through  a  point  P,  distant  d  from  the  ground. 

The  first  step  is  to  locate  the  equivalent  line  charge  in  the 
original  conductor  of  radius  R  and  distance  h  from  ground. 
We  have, 

/ii2  =  h2  -  R2, 
.".  hi  =  Vh2  -  R2  (1) 

from  A,  the 
(2) 

(3) 
(4) 


FIG.  107. 


h2  -  R2 

To  find  the  radius  of  a  circle  whose  center  is 
location  of  the  equivalent  line  charge,  we  have, 


But  from  the  figure  we  have, 

hi  +  ttl  =  fa  +  d 
.*.  «i  =  RI  +  d  —  hi. 


CYLINDRICAL  CONDUCTORS 


225 


Substituting  (4)  in  (2), 
(Ri  +  d-  /ii)  (2/i!  + 


-to 


-  hi) 


2d 


(5) 


The  potential  of  the  circle  of  radius  Ri,  which  goes  through  the 
point,  P,  is: 

Fl  =  2Q  log  ^  =  2Q  log  *' 


But  V,  the  potential  of  the  conductor,  is: 


log 


Knowing  the  radius  from  (5),  and  the  center  is  Ri  +  d  above 
ground,  the  equipotential  surface  through  P  can  be  drawn,  and 
the  potential  of  that  surface  is  given  by  (6). 

Potential  of  a  Cylinder  due  to  External  Charges. — In  order 
to  determine  the  potential  due  to  a  number  of  charged  cylindrical 
conductors,  it  is  necessary  to  calculate  the  potential  of  one 
cylinder  due  to  charges  on  other  cylinders  placed  in  the  vicinity. 


FIG.  108. 

Consider  a  line  charge  Q  at  B  in  Fig.  108  and  determine  the 
average  potential  due  to  Q  on  a  non-conductive  cylinder  A.  The 
potential  at  P  is,  as  has  been  shown  : 


V  =  2Q  log  -> 


but  from  the  triangle  OPB, 
r2  =  d2  -  C2drl  cos  0 


-    -~  cos 


ELECTRICAL  ENGINEERING 

r  =  d  V/c2  +  1—  2k  cos  <f>, 


226 

or, 
where 


Thus  the  average  potential  of  A  is 


=  TT  I 

*TrjQ 


log 


r 

Jo 


log  (1 


-       cos 


-       cos 


/"2r 


where 


2Q  \oghi-2Q\ogd-          I      log  (a-b  cos 
a  =  1  +  fc2 


and,  b  =  2k. 

Evaluating   the    definite    integral    (see    PIERCE'S    "Table    of 
Integrals")  we  find  that  the  last  term  is  zero. 


Thus, 


VA  =  2Q(log  h  -  log  d)  =  2Q  log  -j1 


(D 


Thus,  the  average  potential  is  independent  upon  the  radius  of 
the  conductor. 

But  equation  (l)has  been  shown  previously  to  be  the  potential 
at  a  point  distant  d  from  a  line  charge  distant  hi  above  ground. 

Thus  to  determine  the  potential  of  a  cylindrical  conductor  A, 
due  to  a  line  charge  at  B  distant  d,  the  diameter  of  the  conductor 
does  not  enter  as  long  as,  with  metallic  conductors,  the  field  can 
be  assumed  not  disturbed  by  the  conductor. 


^-^ 

?£>                   / 

>\ 

I                  dz 

Wa 

_X 

h* 

\ 

FIG.  109. 

Referring  to  Fig.  109, 

The  potential  of  A  due  to  B  is: 


log     - 


CYLINDRICAL  CONDUCTORS  227 

The  potential  of  A  due  to  C  is : 

F2  =  2Q2  log  |- 

.'.  V  =  Fi  -}-  F2  =  2Qi  log  ^  +  2Q2  log  ^ 

ai  a2 

Lines  of  Force  between  Parallel  Cylinders. — Let  s-s  (Fig. 
110)  be  a  part  of  a  line  of  force,  and  N-N  a  line  at  right  angles 
to  it.  Thus  the  projection  of  G\  on  the  normal  is  Gi1  =  Gi  cos  a, 


where  G\  is  the  intensity  at  P  due  to  the  line  charge  at  A .  Simi- 
larly the  projection  of  (72  on  the  normal  is  G2  cos  0.  The  sum 
of  the  projections  must  be  zero,  since  N-N  is  perpendicular 
to  the  line  of  force. 


cos  a  -f  G2  cos  |8  =  0 


(1) 


But 


and 


cos 


cos  a  = 


Similarly, 


L      ds 
-2Q 


TT 

as 


15 


228  ELECTRICAL  ENGINEERING 

Substituting  in  (1), 

del  +  de»  =  o 

or  0i  +  62  =  constant. 

This  equation  represents  a  family  of  circles  through  A  and  B, 
with  center  on  the  line  0-0. 

Construction  of  Lines  of  Force. — Referring  to  Fig.  Ill,  as  P 
is  in  the  center  line, 


n 

=  Gl  cos 


or, 


FIG.  111. 

Knowing  the  values  of  x  and  the  fixed  points,  A  and  B,  the 
lines  of  force,  being  circles,  can  be  readily  constructed. 

Problem.  —  Draw  equipotential  surfaces  around  a  line  charge 
placed  10  cm.  above  the  neutral  plane,  when  the  charge  is  1 
electro-static  unit  per  centimeter  of  conductor. 

Find  the  radius  of  the  conductor  containing  the  line  charge 
whose  potential  is  2000  volts.  Draw  surfaces  corresponding 
to  400,  800,  1200  and  1600  volts. 

Draw  lines  of  force  whose  intensities  at  the  neutral  plane  are 
120,  110,  100,  90  and  80  volts  per  centimeter. 

Solutions.  — 

First.  —  Radius  of  conductor:  Since  2000  volts  corresponds  to 
6.67  electro-static  units,  we  have: 


6.67  =  2Q  log 


hi  +  Vhi2  +  R2       0  .      10  +  \XI66~+  fl2 
-  =  2  log  - 


R 


R 


..           10  +  VlOO  +^        n^Q/ix/QQ        1  AA^ 
.  .  logio ^ =  0.434  X  3.3  =  1.445. 


R 


.   10  +  V 100  +  R2 
R 


=  28.05     .*.  R  =  0.72  cm. 


By  a  similar  process  the  radii  corresponding  to  1600,  1200,  800 
and  400  volts  are  found. 


CYLINDRICAL  CONDUCTORS 


229 


These  being  calculated,  the  corresponding  values  of  A,  the 
distances  from  the  neutral  plane,  are  found  by  the  relation 
h  =  VV  +  R2- 


Second.  —  To  find  the  intersection  between  the  neutral  plane 
and  the  line  of  force  of  intensity  100  volts  per  centimeter  or  0.333 
electro-static  units,  we  have: 


-  1  =  10 


).333  X  10 


-1  =  10X0.447  =  4.47  cm. 


Capacity  of  Two  Cylindrical  Conductors,  when  the  Effect  of 
the  Proximity  of  the  Earth  is  Considered. — Consider,  for  the 
sake  of  simplicity,  the  case  of  two  cylinders  of  equal  radii,  and 
charges  Q  and  Qi  respectively. 


FIG.  112. 


Referring  to  Fig.  112,  it  has  been  shown  that  the  potential 
of  A  due  to  its  own  charge,  Q,  and  the  charge  on  its  image,  A'  is: 


2Qlog 


R 


(1) 


It  has  also  been  shown  that  the  potential  of  A  due  to  the  Qi,  on 
conductor  B  is: 

V,  =  2Q1  log  ^  (2) 

Similarly,  the  potential  of  A  due  to  the  image  of  B  is  : 

V,  =  2Qi  log  -1  (3) 


230  ELECTRICAL  ENGINEERING 

Thus  the  total  effect  of  conductor  B  on  A  is: 


Vz  +  V,  =  2Ql  log  ~  (4) 


And  the  resultant  potential  of  A  is: 

" 


VA  =-  Vl  +  F2  +  V3  =  2Q  log 
Similarly, 

^-^log^f^-H*]**  (6) 

Special  Cases.  —  Two  wires  in  parallel  at  same  distance  from 
ground. 

Thus  h  =  H,  Q  =  QL     .'.  VA  =  FB  ==  7. 


Thus  the  capacity  per  centimeter  of  each  wire  is: 

Q  l 


2  log 




[I" 


R  J 

and  the  capacity  of  the  two  wires  taken  together,  is: 

c°  =  -  -]>    >  +  -i/M^fiT  (8) 

\d       h  +  vft2  —  #2 

108  U'        ~R~    "J 

In  the  case  of  a  transmission  line,  ft  is  large  compared  with 
R,  and  df  is  approximately  2ft. 

1  1 


It  has  been  shown  that  the  capacity  of  a  single  wire  to  neutral 
is: 

approximately.       (10) 


2  log  "   '     -  "        "        -  '      2/i 

JLl/  Xl/ 

Thus  the  proximity  of  the  other  wire  has  reduced  the  capacity 
of  each  wire,  so  that  the  combined  capacity  of  the  two  in  parallel 
is  usually  not  more  than  25  to  30  per  cent,  greater  than  that  of 
a  single  wire. 


CYLINDRICAL  CONDUCTORS  231 

As  an  instance,  let  R  =  0.5  cm.,  h  =  1,000  cm.,  and  d  =  20  cm. 


2000 
(log  --  +  log 


=  0.0388  cm.  per  centimeter; 


.*.  2Ci  =  0.0776  cm.  per  centimeter 
and  the  capacity  of  one  single  wire  alone  is 

C  """2000  =  0-0603  cm.  per  centimeter. 

21°six<r 

The  capacity  of  the  double  wire  is  thus  only  28.7  per  cent. 
greater  than  that  of  a  single  wire. 

Second.  —  Assume  now  that  wire  A'  forms  the  return  for  A,  so 
that  the  charge  on  A  is  Q  and  that  on  B  is  —  Q. 

From  equation  (5), 


2Q  log    -     -  -      approximately; 


and  rt  ~  .      (d'     R 

VB  =  2Q  log        •          approximately. 


-  V    = 


If  the  effect  of  the  ground  has  been  neglected,  then,  as  has  been 
shown,  the  capacity  between  the  two  wires  would  have  been 
approximately  : 

c  =      " 


Comparing  equations  (11)  and  (12),  it  is  evident  that  since  -«• 

is  always  smaller,   but  usually  only  very  little  smaller  than 
unity,  C|  is  slightly  greater  than  C. 

The  proximity  of  the  ground  has  thus  slightly  increased  the 
capacity  between  wires.  In  transmission  lines,  the  increase 
amounts  usually  to  less  than  1  or  2  per  cent. 


CHAPTER  XX 

MUTUAL  AND  SELF-INDUCTION  OF  ELECTRO  -STATIC 
CHARGES  OR  FLUXES—  MAXWELL'S  COEFFICIENTS 

If  among  a  number  of  conductors  say  No.  1,  No.  2,  etc.,  a 
particular  one,  say  No.  1,  is  given  a  charge  qi,  so  that  its  potential 
is  Vi,  and  if  all  other  conductors  are  connected  tpo  ground,  that 
is,  are  at  zero  potential,  then, 


where  Ki.\  (with  its  two  indices)  is  called  the  coefficient  of 
self-induction  of  electrostatic  charge,  and  is,  as  seen,  the  capacity 
of  No.  1  due  to  its  own  charge  q\,  when  all  other  conductors  are 
at  zero  potential. 

Obviously  while  the  potential  of  the  other  conductors  is  zero, 
each  has  a  certain  part  of  the  induced  negative  charge  corre- 
sponding to  qi  on  No.  1. 

The  charge  on  No.  2,  for  instance,  is  of  course  proportional  to 
the  potential  of  No.  1  and  is  written: 


Similarly, 

#3  =  Ka.iVi,  <?4  =  Kt.iVit  etc. 

KZ.IJ  KS.I,  etc.,  are  called  the  coefficients  of  mutual  induction. 
Since  Vi  is  positive,  #2  must  be  negative,  therefore,  K2.i,  or  in 
general,  K  with  two  different  indices,  is  always  negative,  while  K 
with  same  indices  is  positive. 

If  instead  of  grounding  all  of  the  conductors  except  No.  1,  we 
now  ground  all  but  No.  2,  and  this  is  given  a  potential  ¥2,  we  get, 
by  a  similar  reasoning, 

q*  =  K2.iVz,     qz  =  #3.2^2,     #4  =  #4.2^2,  and,  qi  =  K^Vz. 
Superimposing  these  conditions,  it  is  readily  concluded,  that, 
if  at  any  time  the  potential  of  No.  1  is  Vi,  that  of  No.  2  is  F2, 
etc. 

232 


MUTUAL  AND  SELF-INDUCTION 


233 


The  following  relation  obtains,  if  Qi,  Q2,  Qs,  etc.,  are  the  total 
charges  on  No.  1,  No.  2,  etc.: 


Qi  = 


A  little  consideration  will  show  that 
KLZ  =  /V2.i,  etc. 

The  applications  of  these  relations  will  be 
illustrated  in  the  case  of  the  two  similar 
overhead  wires  (Fig.  113).  The  immediate 
problem  being  to  determine  the  values  of 
KI.H  Kz.2  and  Ki.%. 

On  account  of  symmetry,  KI.I  =  Kz.zy 
thus  we  have  really  only  two  unknown 
quantities,  namely,  KI.I  and  KI.Z. 

To  determine  them,  give  two  equal 
charges  +Q  to  the  conductors,  then  Vi  —  F2. 


(1) 


20—  x? 


,, 


From  (1),  &  = 


i  +  K^V,  = 
=         =  C  = 


+ 


d'     2h\ 
t 


I        2 
log  (d  '  R 


[See  (9)  in  the  previous  article.] 

Now  give  one  conductor  a  charge  +Q  and  the  other  a  charge 
—  Q,  so  that  the  potential  of  No.  1  is  Vi  and  that  of  No.  2  is 
-Fi,  then  from  (1),  Qi  =  Vi  (K^  -  K^), 


=  cf  - 


2  log     ' 


(3) 


[See  (11)  in  the  previous  article.] 

From  these  equations  it  follows  that : 

lo     (—} 
KI.I  =  -  5  ™' 


'd     2h\         id'     2h\ 
^d'     R/         \d      R/ 


log    " 


r  -  r 


234  ELECTRICAL  ENGINEERING 

and,  ,      2/i 

#1.1  °g  R       C  +  C' 

#1.2  ,      d'        C  -  C' 


Numerical  application,  Fig.  113: 

Let  R  =  0.5  cm. 

h  =  1000  cm. 
d  =  20  cm. 
/.  d'  =  2000  cm. 

.'•  7^  =  4000, 

?  =  100' 

and>  I  =,  o.oi. 

•'*  C   =  2  log  400,000  =  0<°388' 

C'  =  s-^  -777  =  0.1352. 
2  log  40 

.'.  KL1  =  0.087, 
Ki  .2  =  -  0.0482, 

f^-1  =  -  1.806. 

A  1.2 

Discussion.  —  To  show  the  application  of  these  coefficients,  the 
following  problems  will  be  considered. 

A.  Compare    the    capacities    between    a    wire    and    ground, 
(a)  when  the  wire  is  alone;  (b)  when  an  adjacent  wire  is  grounded. 

B.  Compare  the  charging  currents  for  the  same  applied  voltage 
between  the  two  conductors  when  the  two  wires  are  insulated, 
and  when  one  is  grounded.     In  the  latter  case,  give  the  relative 
proportions  of  the  current  in  the  grounded  wire  and  in  the  ground 
itself. 

The  numerical  case  will  be:  R  =  0.5  cm.; 

h  =  1000cm.; 
and,  d  =  20  cm. 

The  problems  will  be  best  solved  by  the  use  of  the  MAXWELL'S 
equations,  viz. 

Qi  =  #i.iFi  +  Kt.iVt  +  Ki.tV*, 

Q2    =    #J.lFi   +  #2.2F2  +  #2.3^3, 

and,  Q3  =  KZ.1V1  + 


MUTUAL  AND  SELF-INDUCTION  235 

In  these  equations,  index  1  refers  to  conductor  No.  1,  index 
2  to  No.  2,  and  index  3  to  the  ground. 

Since  the  potential  of  No.  3  is  zero  and  since  we  assume  two 
similar  and  similarly  placed  conductors, 

V3  =  0,  KM  =  #2.2  and  Klft  =  K,.3. 

/.  Qi  =  K1.1V1  +  #!.2F2  (7) 

Q*  =  #i.2F!  +  #i.iF2  (8) 

and  Q3  =  #i.3Fi  +  K^V*  (9) 

Case  A.  —  (a)  It  has  been  shown  that  with  a  single  conductor 
suspended  above  ground,  the  capacity  is: 

C  =  --  ^r  =  0.0601  cm.  per  cm.  (10) 

r»    i  •""' 

21ogfl 
Thus  if  V  is  its  potential  the  charging  current  is: 

ri  =  0.0601  ^ 

(6)  since  No.  2  is  grounded,  T2  =  0. 

Thus  from   (7),  Qi  =  ^i.iFi     .'.  capacity  =  KI.I  =  0.087,  and 

ri  =  0.087  ^ 

The  capacity  of  wire  No.  1  is  increased  45  per  cent,  by  the 
proximity  of  the  grounded  adjacent  wire  No.  2. 
Case  B.  —  Under  normal  conditions, 

Qz  =  -  Qi  and  KI.I  =  K2.z, 
.'.    Qi  =  tfi.iFi  +  X!.2F2, 


Thus  the  capacity  between  the  conductors  is: 
C  =  g"  ~  g"    =  0.0676. 

£i 

If  FI  —  F2  =  F,  and  if  z'i  is  the  current  in  conductor  No.  1, 
then 

dV  dV  dV 

il  =  C^  =  y2  (#!.!  -  #,.,)  ^-  =  0.0676^- 


236  ELECTRICAL  ENGINEERING 

When  No.  2  is  grounded,  Vz  =  0. 

.'.  Qi  =  Ki.iVi  =  Ki.iV,  thus  the  capacity,  C"  = 


=  1.285. 


*1          JV.1.1  ~  A- 1.2 

The  charging  current  in  conductor  No.   1  is  increased  28.5 
per  cent,  by  the  proximity  of  the  adjacent  grounded  wire. 
The  charge  in  conductor  No.  2  is: 

Qz  =  K2.iVi  -f  #2.2     Vz  =  Ki.2V,  since  72  =  0. 
But    Xi.t  =  -  0.135  +  0.087  =  --  0.048 

Thus      ^2=  -0.048  f 

The  current  carried  in  the  ground  is  obviously 
-  23  =  (0.087  -  0.048)  -^ 

.:  ,-,--  0.039  £ 

If  the  current  in  No.  1  after  grounding  No.  2,  is  taken  as  1  amp., 
then  wire  No.  1  carries  1  amp.,  No.  2,  0.554  amp.  and  the 
ground,  0.446  amp. 

Problem. — Assume  three  similar  horizontal  conductors  of 
R  =  0.5,  h  =  1000  and  d  =  20. 

Give  the  relative  values  of  the  charging  current  between  No.  1 
and  No.  3  if  No.  2  is  indulated,  and  if  it  is  grounded.  Also  give 
the  charging  current  if  No.  2  is  removed  entirely.  Consider 
the  current  in  the  last  case  to  be  unity. 


CHAPTER  XXI 
TWO-CONDUCTOR  CABLE 

Since  the  conductors  as  well  as  the  lead  covering  are  of  metal, 
the  surfaces  of  each  are  equipotential  surfaces.  In  order  to 
simplify  the  calculations  it  is  desirable  to  substitute  for  the  sheath 
and  each  conductor  a  system  of  conductors,  i.e.,  the  conductor, 
and  its  image,  which  will  give  the  same  distribution  of  potential. 

Consider  first  the  system  of  Fig.  114  consisting  of  A,  its  image 
A'  and  the  lead  sheath.  It  is  necessary  to  determine  the  position 
of  the  line  charges  at  distance  hi  from  the  neutral  plane,  so  that 
the  conductor  A  and  the  sheath  are  equipotential  surfaces. 

From  what  has  been  shown  previously,  it  is  evident  that  the 
following  relations  exist: 


and 


hi2  =  h2  —  r2,  when  considering  the  conductor; 

hi2  =  (h  +  a)2  —  r*i2,  when  considering  the  sheath. 

i2  -  r2  -  a2 


Having  determined  h  from  (1), 

hi  is  determined,  as  hi  =  \/h2  —  r2 


(1) 


(2) 


FIG.  114. 

Referring  to  Fig.  114,  it  is  evident  that  the  potential  of  A  is 
due  to  its  own  charge  and  the  charge  on  its  image,  and  the  charges 
on  B  and  its  image. 

TVf) 

It  is  also  recollected  that  the  latter  potential  is:  2Q  log  =» 

mp 

237 


238  ELECTRICAL  ENGINEERING 

if  we  neglect  the  shortening  of  the  lines  of  force  from  m  to  p  in 
going  through  conductor  B,  where  np  is  the  distance  between  the 
line  charge  in  B  and  the  center  of  A,  and  mp  is  the  distance  be- 
tween the  line  charge  in  B'  and  the  center  of  A. 

.'.  np  —  2a  -f-  h  —  hi, 
and  mp  =  2a  -f-  h  +  hi. 


21      r'  +  VW-r'^a  +  h-h! 


to  neutral   (3) 
i"  —  r*     za  -f  AI  —  Aii  \ 

Approximation. — Frequently,  in  fact  almost  always,  the  follow- 
ing approximation  can  be  made : 

h  =  hi. 

(4) 


2  log  - "' 


r  a 

If  furthermore  r2  is  small  compared  with  ri2  —  a2,  and  is  small 
compared  with  A2,  then, 

h  -4-  A//?2  —  r2       2h       ri2  —  a2 


- 

/I    — 


. 

2a  r  r  ar 


thus,      c  =          _^ —  __  ^  the  capacjty  to  neutral        (5) 

2  log  ( —  •     10   , 
\r      r^  + 

Thus,  the  capacity  between  the  two  conductors  is  approximately 

C"  /aa'r..        -  (6) 

4  log  (7-  "  ^rf 

or,  in  microfarads  per  1000  ft.  of  cable, 

(7) 


, 

log 


l° 


To  determine  the  capacity  of  the  two  conductors  in  parallel 
against  the  sheath,  the  two  conductors  are  given  positive  charges, 
+Q,  and  hence  the  charges  on  the  images  are  —Q. 


TWO-CONDUCTOR  CABLE  239 

• 
The  potential  of  A  due  to  its  own  charge  and  the  charge  on 

its  image  is: 

V'A  -  2Q  log  *±X^Ei 

The  potential  of  A  due  to  the  charges  on  B  and  its  image  is: 


or,  using  the  same  approximations  as  before, 

(9) 


The  potential  of  the  sheath,  if  insulated,  due  to  the  charges 
in  A  and  its  image  is: 

V.  =  2Q  log  *' 


Similarly,  due  to  B  and  its  image  is : 
y".  =  2Qlog^-± 


r\ 


.-.  F.  =  4Q  log  (10) 

Tl 

Using  the  same  approximations  as  before, 

V8  =-  4Q  log  ^  (11) 

Thus  the  potential  difference  between  the  sheath  and  either 
of  the  conductors  (when  they  are  connected  in  parallel)  is 
approximately  : 


V    -VA-A.  =  2Qlog-~-  -  2Qlog 

(12) 


Thus  the  total  capacity  between  the  two  conductors  in  parallel 
and  the  sheath  is: 


log 


240 


ELECTRICAL  ENGINEERING 


In  connection  with  this  it  may  be  of  interest  to  determine  the 
capacity  between  the  conductor  and  the  sheath  in  a  single  con- 
ductor, eccentric  cable,  Fig.  115. 

The  potential  of  A  due  to  its  own  charge  and  the  charge  on 
its  image  is: 

n  +  VV  +  r2 


VA  =  2Qlog 


FIG.  115. 

The  potential  of  the  sheath  due  to  the  charge  on  A  and  its 
image  is: 

78  =  2Q  log 


hi  +  Vhi2  + 


.'.  C  = 


II 


(13) 


Denoting  the  conductor  A  with  1,  B  with  2  and  the  sheath 
with  3,  the  values  of  KI.I,  KI.%  and  KI.S,  are  respectively  identical 
with  K2.2,  K2.i  and  ^2.3.  To  determine  them  we  have 


and,  Q3  =  KZ.1V1  +  ^3.272  +  #3.373. 

If  we  are  concerned  with  the  distribution  of  currents  in  the 
conductors  and  lead  sheath,  it  is  convenient  to  consider  the  sheath 
grounded,  that  is,  73  =  0. 


and, 


TWO-CONDUCTOR  CABLE  241 

If  then  FI  =  F2  =  F,  that  is,  if  both  conductors  are  given 
the  same  positive  charge,  then 

Qi  «  V(#1;1  +  #,.2)     /.  C  =  #1.1  +  #1.2; 
but  C  has  been  determined  in  (12)  which  gives, 

(14) 


If  the  two  conductors  have  potentials  FI  and  —  FI,  respectively, 
then: 

Qi  =  #1.1^  +  #1>2F2  =  F^L!  -  #i.2), 
.'.  C  =  #1.1  ~  #1.2- 

This  capacity  has  been  given  in  (5),  which  is: 

1 


C  = 


2a 


r!2  +  a 

From  equations  (14)  and  (15)  the  values  of  KI.\  and  KI.Z  are 
readily  obtained. 

Consider  finally  that  when  the  two  conductors  are  in  parallel, 
that  is,  at  the  same  potential  and  the  charging  current  returns 
over  the  grounded  sheath,  we  have, 

Qi  +  Q2  +  Q3  =  0,  and  Vi  =  F2  =  V. 

.'.  (Ki.i    +   Ki.2   +  KM   +  #2.2   +  #3.1   +   #3.2)    F    =    0, 

or,  2#!.!  +  2Ki.2  +  2#!.3  =  0. 

.*.  #1.3  =  -  (#1.1  +  #1.2)  (16) 

Problems.  —  Find  the  charging  current  under  the  conditions 
shown  in  Figs.  116-120,  when  rt  =  4r;  a  =  2r;  .'.  h  =  2.75r; 
hi  =  2.55r;  KI.I  +  #1.2  =  0.4;  #1.1  -  KI.I  =  0.57  .'.  #1.1  = 
0.485;  #1.2  =  0.085  and  #1.3  =  —  0.4  (using  no  approximations). 
(a)  (Fig.  116.)  F!  =  F2,  F3  =  0. 


Q2  =  Fi(#i.2  -f-  #2.2). 
.'.  ii  —  (#1.1  +  #1.2)  -jj£, 

dV 
and  iz  =  (#1.1  +  #1.2)  ~9 


242  ELECTRICAL  ENGINEERING 

the  total  charging  current  is 

1  "*"     l'^  ~dt   '      '     dt' 


(6)  (Fig.  117.) 


'i.iVi  +  Xi.2F2, 


.-.<h - 


i.i  -  K 


1.2 


dV 

:  ' 


(c)  (Fig.  118.)     72  =     -  Fi,  F  =  27!. 


o 


FIG.  116.  FIG.  117. 

S 


FIG.  118. 


FIG.  119. 
(d)   (Fig.  119.)      F2  =  0. 


FIG.  120. 


=  0.485       . 


dV 


TWO-CONDUCTOR  CABLE 


243 


(e)  (Fig.  120.)  Charging  current  in  the  eccentric  cable:  Since 
the  shortening  of  the  lines  of  force  in  going  through  a  conductor 
is  neglected,  when  the  formulas  were  developed;  so  the  solution 
of  case  (6)  is  a  solution  of  case  (e), 

dV 


0.47 


~df 


Three-phase  Cable.  (Fig.  121.) — -The  location  of  the  inverse 
points  is  determined  as  in  the  case  of  the  two-conductor  cable. 
Thus  h2  =  h\2  +  r2,  when  considering  A  and  A' '; 


FIG.  121. 
and  (h  -f-  a)2  =  hi2  +  ^i2,  when  considering  the  sheath. 

•'•  h  -  r'   ~2a~  -' 


and  At  = 

Thus  /ii  is  known. 

Let,  at  a  given  instant,  the  charges  on  A,  B  and  C  be  QA, 

and  Qc  respectively. 

The  potential  of  A  due  to  the  charges  on  A  and  A'  is: 

V  =  2QA  log  - 

16 


(1) 


244  ELECTRICAL  ENGINEERING 

The  potential  of  A  due  to  the  charges  on  B  and  B'  is  : 

V  =  2QB  log  £ 
y 

The  potential  of  A  due  to  the  charges  on  C  and  C"  is: 
V  =  2Qc\0g~- 

\J 


/?t    -4-    -V/)    2      I      r2  ~ 

.'.  VA  =  2QA  log  -        ^         •  +  2(QB  +  <?.)  log  ?      (2) 

x  is  the  distance  between  the  inverse  point  of  £'  and  the 
center  of  A,  and  ?/  is  the  distance  between  the  inverse  point  of 
B  and  the  center  of  A. 

With  a  very  slight  approximation,  the  distance  y  may  be 
counted  between  the  respective  centers,  thus, 

y  =  a  V3  (3) 

and, 

x*  =  (h  +  hi  +  a)2  +  a2  -  2a(h  +  fci  +  a)  cos  120°       (4) 

leth  +  hi  +  a  =  D  then, 

x2  =  D2  +  a2  +  aZ)  (5) 

It  has  been  shown  previously,  that  since  the  sheath  is  an  equi- 
potential  surface, 

(a  +  a)D  =  n2  =  (a  +  h  -  /ii)  D  =  n2, 

.'.  D  =  -  T-Y^-  (6) 

a  +  h  —  hi 

Approximations  based  on  the  usual  conditions,  that  h  is  very 
nearly  the  same  as  hi  and  r2  is  small  compared  with  h2. 
Referring  to  equation  (2), 

hi  +  Vhi2  +  r2       2/i       D  -  a 


r  r  r 

2 


(7) 


Referring  to  (6),          n  _          »V          .  » 
U  ~  a  +  h-h,"   a 


• 


a 
Referring  to  (3),  y  =  a  V3  (10) 

The  potential  of  A  due  to  charges  on  A  and  A'  is  =  2QA  log  —  -  — 
The  potential  of  B  due  to  charges  on  A  and  A'  is  =  2QA  log  —  -/=• 


TWO-CONDUCTOR  CABLE  245 

The  potential  of  C  due  to  charges  on  A  and  A'  is  =  2QA  log  - 
The  potential  of  A  due  to  charges  on  B  and  B'  is  =  2QB  log 


D—  a 
The  potential  of  5  due  to  charges  on  B  and  5'  is  =  2QB  log  - 

/v» 

The  potential  of  C  due  to  charges  on  B  and  B'  is  =  2Q#  log 


The  potential  of  A  due  to  charges  on  C  and  C'  is  =  2QC  log  -  -r--- 
The  potential  of  B  due  to  charges  on  C  and  C'  is  =  2QC  log 


The  potential  of  C  due  to  charges  on  C  and  C'  is  =  2QC  log  -  --  -• 

Since  I,VA  +  2FB  +  27C  =  0  in  a  three-phase  system,  we  get, 
by  adding  all  the  equations  given  above, 


2  log  ~        =  0 

or,  QA  +QB  +  Qc  =  0  (11) 

which  really  needed  no  proof  from  our  knowledge  of  the  char- 
acteristics of  the  three-phase  system. 

From  (11)  follows  that  QB  +  Qc  =  -  QA- 

/.  VA  =  2QA  log  ^         -  2QA  log  ^ 

(12) 


.',   the  capacity  of  A  to  ground  or  neutral  is: 


2  log 


ryVi4  +  a4  +  r!2  a2/ 


or  in  microfarads  per  1000  ft.  of  cable,  to  neutral, 

0.00736 


(13) 


(14) 


+  a4  +  r    a 


2 


246  ELECTRICAL  ENGINEERING 

In  order  to  determine  MAXWELL'S  coefficients,  by  symmetry, 
we  have: 


and,   KI.Q  =  K2.Q  =  K3.0. 

where  index  o  represents  the  sheath.     It  is  necessary  to  calculate 
the  capacity  between  all  three  conductors  and  the  sheath. 

Assume  thus  that  the  three  conductors  are  given  the  same 
positive  charge  Q,  and  that  the  images  therefore  have  charges 
—  Q.  The  potential  of  A  due  to  the  three  charges  is  evidently 

VA  ,  2Q  lo,  +  2Q  log  +  20  1.8 


The  potential  of  the  sheath  is  due  to  the  charges  in  the  three 
conductors  and  since  the  sheath  is  symmetrical  with  reference 
to  each  conductor  and  its  image,  we  get: 


or,  from  the  illustration,  neglecting  a, 

Fo-3X2Qlog^  =  2 
or  since  aD  =  rf, 


log  (16) 


.'.  C  =  -  —  i,  between  a  conductor  and  the  sheath. 

ri6  -  a6' 

21og  377W 

It  is  now  possible  to  determine  the  values  of  Kimi,  K^  and 
KI.Q.     Assume  that  the  sheath  is  grounded,  that  is,  V0  =  0. 

/.  Qi  =  KlmlVi  +  K^V*  +  K^V*  =  K^Vt  +  ^i.2(F2  +  V,). 
Since  Vi  +  72  +  F3  =  0,     F2  +  F3  =  --  yt. 


TWO-CONDUCTOR  CABLE 

1 


It  follows  then  from  (13)  that, 

#1.1    —    #1  2    =    - 


2  log 


«\/3  (r!2  -  a2) 


247 


(19) 


+  a4  +  ri2  a2/ 

Considering  next  the   case   when   all   three   conductors  have 
the  same  charge,  then: 

Qi  =  #i.i7i  +  #i.272  +  #i.878  =  (#1.1  +  #1.2  +  #1.3) 7i  = 


From  equation  (18)  it  follows  that, 

Kl      Q  if 
1.1   -f-  ^^-1.2    - 


1 


(6    n6 
'i^r 


(20) 


From  (19)  and  (20)  KI.I  and  Ki2  can  be  solved. 

To  determine  K\mQ,  assume  that  not  only  the  three  conductors 
but  also  the  sheath  is  given  a  potential  V,  in  which  case  the 
charge  is  confined  to  the  sheath  only.  Then: 

0  =  (#!.!  +  Xi.2  +  tfi.3  +  KwW,  .'•  ^1.1  +  2X1>2  +  ^1.0  =  0; 
0  =  (X2.i  +  X2.2  +  K2.3  +  X2.0) 7,  .'.  #1.1  +  2X1.2  +  XLO  =  0; 
0  =  (X8.i  +  #3.2  +  K3.3  +  X8.o)7,  /.  #1.1  +  2#!.2  +  Xi.0  =  0; 

any  one  of  these  equations  gives: 

#1.0  =     -  (#1.1  +  2Xi.2)  (21) 

Thus  XI.Q  is  determined. 

Problem. — Verify  the  equations  of  the  charging  current  under 
the  conditions  given  below  (Figs.  122-130)  and  apply  the  follow- 
ing numerical  values: 

TI  =  4r,         a  =  2r. 


(Fig.  122)      i  = 
(Fig.  123)     i  =  2 


FIG.  123. 


;  ^     -          0<826- 

Xi.i  /  dt  dt 


248 


ELECTRICAL  ENGINEERING 


(Fig.  124)    i  =  3(/M.i  +  2/M.j)  —^  =  0.903 
Ki.i2  -  #i.22  dV 


(Fig.  125)     t 


dt 

M*  dV 
dt 


.      _  ZKi.z* 

to  —  ~Tr~       —  A.  1.1 

-IV  1.1 

Si  =  0 


FIG.  128. 
(Fig.  126)    ti  = 


FIG.  129. 


K     — 
dV 

l*  =  K™~di 

i          K      dV 

13  ~  Kl-*~di 


=       l'° 


St  =  0 


dt 
(Fig.  127)      i  =  2(£i.i  +  Klm- 

~K"  K"        A 

(Fig.  12J 
(Fig.  129)      i 


dV 


i  = 


0.608 


=  -  0.418 

=  0. 


dV 
dt' 


FIG.  127. 


v      v 
FIG.  130. 


=  0. 
=  1.488 


dV 
dt' 

~dt' 


(Fig.  130)  Three-phase:  z  =  (Klfl  -  K^)  -     =  0.744 


CHAPTER  XXII 

THE  ELECTROSTATIC  EFFECT  OF  A  THREE-PHASE 
LINE  ON  AN  ADJACENT  WIRE  OR  WIRES 

The  potential  of  the  wire  W,  Fig.  131,  due  to  A,  B,  and  C 
and  their  images  is  obviously: 

V  =  2QA  log  g  +  2QB  g  +  2QC  log  g 

2QBb,  +  2Qcd  (1) 


where  .      r2 

01  =  log  —  > 

61  =  log  p 

and,  ,      r6 

Cl  =  log  -• 

If  C  is  the  average  capacity  of  the  three  lines  against  neutral. 
then:  QA  =  Ce\,  QB  =  Cez,  and  Qc  =  Ces,  where  e\t  e2  and  es 
are  the  instantaneous  values  of  the  Y  voltages. 


e26i  -f  e8ci) 
2CE[ai  sin  6>  +  bi  sin  (^  +  120°)  +  d  sin  (^  +  240°)]. 


-  60  +  b^  -  cO  +  Ci(Cl  -  ai)  (2) 


where  £7  is  the  maximum  value  of  the  Y  voltage,  that  is,  of  the 
voltage  to  neutral. 

To  determine  the  average  capacity  of  the  three  wires:  The 
potential  of  A,  Fig.  132,  due  to  its  own  and  the  other  charges  is 
evidently, 

VA  =  2QA  log  ^  +  2QB  log  g  +  2QC  log  ||- 

If  the  average  value  of  R2,  R*  and  RQ  is  #!,  and  the  average 

249 


250 


ELECTRICAL  ENGINEERING 


values  of  R3  and  R$  is  D,  then  the  potential  of  A  can  be  reason- 
ably well  expressed  as: 

VA  =  2QA  log  y  +  2QB  log  ^  +  2QC  log  ~ 

(3) 


WA  log  -1  +  2  (Q*  +  Qc)  log  ^ 

7; 


r  =  Radius  of 
Conductor 


FIG.  131. 
But  QB  +  Qc  =  -QA,  thus 

7A  =  2QA  log  (~  •  j^   = 

1 


FIG.  132. 


/.  C  = 


where  D  is  the  average  distance  between  the  conductors. 


E 


'.'•      Vmax.    =    T?\/<ll(Q>l  —  61)  +  61(61  —  Ci)  +  Ci(Ci  - 


(4) 


(5) 


Problem. — Prove  that  the  maximum  value  of  the  induced 
potential  on  a  telegraph  wire  placed  under  a  three-phase  trans- 
mission line  of  100,000  volts  (effective)  between  the  lines  is 


A  THREE-PHASE  LINE  251 

approximately  3100  volts,  when  H  =  average  height  of  trans- 
mission wires  above  ground  =  1500  cm.,  D  =  300  cm.,  and 
r  =  0.5  cm.  The  telegraph  wire  is  800  cm.  above  the  ground, 
and  50  cm.  to  the  left  of  the  center  line  of  the  pole. 

It  is  seen  that  when  the  three-phase  line  is  operating  under 
normal  conditions,  the  voltage  induced  in  an  adjacent  wire  is 
only  a  few  per  cent.,  in  this  case  only  3  per  cent,  of  the  line 
voltage.  If,  however,  one  of  the  three-phase  lines  is  grounded, 
so  that  the  system  is  unbalanced  electrostatically,  then  very 
considerable  voltage  is  induced  as  will  be  shown. 

If    ei  =  E  sin  e, 

e2  =  E  sin  (e  +  120°), 
and,   63  =  E  sin  (0  +  240°) 

are  the  Y  voltages  or  phase  voltages, 

then  it  is  well  known  that  the  line  voltages  are : 

V1  -  F3  =  EV3  sin  (6  +  30°), 
and,  V2  -  F3  =  EV3  sin  (e  +  90°). 

Therefore,  if  phase  No.  3  is  grounded  or  at  zero  potential, 
then  we  have  the  relation  between  the  line  voltage  as  shown  in 
Fig.  133.  The  line  voltages  differ  60°  in  time  phase,  when  one 
phase  is  grounded. 

For  the  sake  of  simplicity,  let : 

Vi  -  73  =  £V3  sin  e  =  Va;    ' 
V2-  V,  =  EV3  sin  (6  +  60°)  =  76; 
73  =  0  =  Vc. 
or,  Va  =  #o  sin  0; 

where  EG  =  E\/3> 
Vb  =  E0  sin  (6  +  60°) ;  FIG.  133. 

Using  MAXWELL'S  equation,  applying  index  e  for  ground, 
and  remembering  that  Vc  =  Ve  =  0,  we  have, 

Qa    =   Ki.iVa  4~  Ki.zVb, 

Qb  =  #2.iF«  +  K*.2Vb, 

Qc    =    #3.lF«   +   #3.2F6, 

and, 

Q.  =  K..iVa  +  Ke.2Vb. 


252  ELECTRICAL  ENGINEERING 

But  KI.I  —  X2.2,  Xi.2  =  Xi.s  and   Ke.i  =  Xe.2  approximately. 
•*•  Qa  —  Xi.iFa  H-  Ki.zVb, 
Qb  =  Xi.s7.  +  Xx.iFi, 
Qc  =  K^Va  +  Xi.2F6, 

and,  Qe  =  Xi.,7.  +  Xi.6F6. 

•'.  Qa  =  Xi.i#0  sin  0  +  #1.2  EQ  sin  (0  +  60) 


Xi.i  +  0.6X1.2)  sin  0  +      -  #0  Xi.2  cos  0  (6) 

sin  6>  +  Xi.i  ^0  sin  (0  +  60°)  =  #0(^1.2  +  0.5Xi.i) 
Sin  0  +  --KLI  cos  0  (7) 


Qc  =  Xi.2(F0  +  Vb)  =  E0Ki  2(1.5  sin  6  H — pr—  cos  6)       (8) 

£t 

and,  Qe   =  Ki.e(Va+  Vb)  =  E0Ki.e(1.5  sin  d  +  -g—  cos  0)        (9) 

Assuming  for  the  present  that  the  values  of  the  MAXWELL'S 
coefficients  are  known,  it  is  then  possible  to  obtain,  in  a  manner 
similar  to  that  used  for  the  balanced  system,  the  potential  of 
the  telegraph  wire. 

While  in  this  case  we  deal  with  four  charges,  the  effect  of  the 
charge  of  the  earth  is  not  felt  at  the  telegraph  wire,  because 
the  earth  may  be  considered  as  an  infinite  cylinder,  enclosing 
all  wires;  thus  the  effect  of  its  charge  on  any  point  inside  it,  re- 
sults in  no  potential.  The  potential  of  the  wire  is  now  readily 
obtained  from  equation  (1).  The  charging  current  in  the  three 
wires  and  the  earth  is  found  from  equations  (6)  to  (9),  remem- 
bering that  0  =  cot. 

.    .      dQa  r  \/3  .       I 

•   •  ^a  =  ~TT   =  L(\CO\   (/Vi   i  H~  U.O/Vi  2)  COS  COl ~ —  Al   2  Sin  001      , 

at  L  2  J 

r  \/3  i 

4  =  EQCO\  (Xi.2  +  O.SXi.i)  cos  cot ~—  XLI  sin  cot     ', 

ie    =    E0OJ\  Xi.2    (1.5   COS   Cot ^~  SHI    CoZ      ', 

le    =    EQCO\  Ki.e    (1.5   COS   Cot ^~  Sm    W0  j 

It  remains  now  to  determine  the  values  of  the  MAXWELL'S 
coefficients. 

Give  each  of  the  three  conductors  the  same  charge  Q,  and 
assume  average  values  of  the  distance  between  the  conductor 


(10) 


A  THREE-PHASE  LINE  253 

and  ground  as  H  and  the  distance  between  conductors  as  D. 
Then  we  have  approximately  the  following  relation: 

OI7  9T/  977 

V.  =  2Q  log  —  +  2Q  log  ^  +  2Q  log  ^  =  2Q  log 


'  '  V 


We  have  also, 

.'.  #1.1  +  2#lt2  -  -  -^  (11) 

2  log  Wr 

Give  now  three-phase  charges  to  the  three  conductors,  then, 

QA      =     ^l.l^a     +     #1.2^6     +     K-l-zVc      =     Ki.iVa     ~     #1.2^6     = 
Va   (#!.!    -   #x.2). 

Thus  #1.1  —  #1.2  is  the  capacity  of  one  of  the  three  lines 
against  the  neutral,  which  has  been  shown  to  be: 

1 


.  .  #1.1  —  #1.2  —  ~         T^-  (12) 

21ogf 

From  (11)  and  (12),  the  numerical  values  of  Kn  and  #i.2 
can  be  determined,  as  well  as  #1.3,  so  that  all  the  coefficients  are 
known. 

It  may  be  of  interest  to  consider  the  problem  from  another 
point  of  view. 

By  grounding  one  conductor,  while  the  potential  difference 
between  the  conductors  is  not  changed,  the  potential  of  the 
system  of  three  conductors  has  been  changed. 

It  should  be  possible,  therefore,  to  calculate  the  charge  Q0, 
which  should  be  given  to  each  conductor,  in  order  that  the 
new  potential  distribution  shall  exist.  The  charge  should  ob- 
viously be  such  that  the  potential  of  C  shall  be  reduced  to  zero. 
Before  grounding,  the  potential  was  +VC,  and  hence  QQ  should 
be  such  as  to  give  C  a  potential  of  —  Vc. 

:.  -Vc  =  2Q0  log  y2  +  2Q0  log  |^  +  2Q0  log  j* 


254  ELECTRICAL  ENGINEERING 


SH3 
=  2Q0  log  -j,  using  the  approximations. 


Since  —Vc  =  —  E  sin  (cot  +  240°),  the  maximum  value  of  the 

TjJ 

charge  is  Q0  =  - 


The  charges  on  the  conductor  A  after  grounding  the  conductor 
C  are  therefore, 

QA  +  Qo  =  E  sin  cot 


Similar  expressions  are  of  course  readily  written  for  the 
charges  on  the  conductors  B  and  C. 

The  potential  of  the  telegraph  line  after  grounding  is  thus, 

V  =  2  [(QA  +  Q0)ai  +  (QB  +  Qo)6i  +  (Qc  +  Qo)CJ. 

By  applying  these  equations  to  the  numerical  example  given 
previously,  it  will  be  found  that  the  induced  potential  of  the 
telegraph  line  will  be  25  per  cent,  of  the  phase  voltage  or  14.5 
per  cent,  of  the  line  voltage.  In  the  case  of  an  insulated  balanced 
system,  it  was  found  about  5  per  cent,  of  the  phase  voltage  or 
about  3  per  cent,  of  the  line  voltage. 

The  Effects  of  a  Grounded  Horizontal  Wire  on  the  Distribution 
of  Electricity  in  the  Atmosphere.  —  It  has  been  observed  that 
frequently  considerable  potential  difference  exists  between 
successive  layers  of  the  atmosphere.  A  potential  gradient  of 
600  volts  per  m.,  or  roughly  200  volts  per  ft.,  is  not  unusual. 

It  is  of  interest  then  to  see  how  much  the  potential  at  a  given 
height  may  be  reduced  by  a  grounded  overhead  line  such  as 
is  used  in  high-potential  transmission  systems. 

Assume  that  the  gradient,  not  far  from  the  earth,  is  2  electro- 
static units  per  m.  (600  volts  per  m.).  It  is  readily  seen  that 
the  distribution  can  be  quite  closely  represented  by  the  effect  of 
a  charged  cylindrical  conductor,  say  300  m.  or  more  above  the 
surface  of  the  earth.  The  conductor  then  represents  whatever 
cause  there  was  for  the  potential  gradient. 

The  charge  per  centimeter  length  of  the  fictitious  conductor 
is  determined  by  the  fact  that  the  potential  at  a  certain  height 


A  THREE-PHASE  LINE 


255 


is  known.  Thus  according  to  the  assumption,  the  potential  at 
15  m.  above  the  ground  is  30  electro-static  units.  Thus  referring 
to  Fig.  134, 

01  C 

V  =  2Q0  log          =  30    .'.   QQ  =  155. 


Suppose  now  that  it  is  desired  to  find  the  change  in  a  grounded 
overhead  wire  of  radius  r  =  0.5  cm.  placed  15  m.  above  ground. 

Since  the  potential  of  A,  Fig.  135,  is  zero,  it  is  evident  that 
the  potential  of  A  due  to  its  own  charge  and  the  charge  on  its 
image  plus  the  potential  of  A  due  to  the  charge  on  the  fictitious 
conductor  and  its  image  must  be  zero. 

nr.  TT     i      TL  n~L 

Thus    2Q  log  - :  +  2Q0  log  JT~  =  0  =  2Q  log  -  •  +  30. 


.'.  Q  =  - 


30 


=  -  1.72  E.S.U. 


T 


P  Abs 


FIG.  134. 


FIG.  135. 


FIG.  136. 


The  potential  at  a  point  P,   Fig.    136,   distant  hi,  from  the 
ground  is  then: 

V  =  2Q0  log  -TT  +  2Q  log  — ,  but  2Q0  log  -^  is,  according  to 

KI  TI  III 

the  first  assumption  of  uniform  gradient,  0.02/ii  (hi  being  given 

in  centimeters). 

Thus  the  potential  of  P  is : 


VP  =  0.02/i!  -  3.44  log  - 


(1) 


The  effect  of  two  ground  wires  A  and  B  on  the  potential  at 
a  point  P  in  the  vicinity  of  the  wires : 


256 


ELECTRICAL  ENGINEERING 


The  potential  of  A  or  B  due  to  the  fictitious  and  the  two  actual 
conductors  and  their  images  must  be  zero. 
The  potential  of  A,  Fig.  137,  is: 

2Q0  log  |4i  +  2<3  loS  y  +  2Q  log  ^  =  0  = 

or  0.02ft  +  2Q  log  y  ^  =  0,  /.  Q  =  — fj^   (2) 


If  the  wires  are  2  m.  apart  and  15  m.  from  ground  then  r4  = 
3010  cm.  and  r3  =  200  cm. 

:.  Q  =  -1.31. 


T 


The  potential  at  a  point  P,  Fig.  138,  is  then: 
Vp  =  0.02^!  -  2.63  log  T-~ 


(3) 


It  will  be  seen  that  by  means  of  a  single  ground  wire  above  a 
transmission  line  the  potential  is  reduced  by  some  30  per  cent., 
and  when  two  ground  wires  are  used  by  some  40  to  50  per  cent., 
and  that  there  is  little  gain  in  using  ground  wires  of  large  diameter. 


CHAPTER  XXIII 


THE  CURL  OF  A  VECTOR 

In  vector  representation,  the  curl  of  a  vector  is  represented 
by  the  cross-product  of  the  differential  operator  V  and  the 
vector.  It  is: 


V  X  R  =  curl  R  = 


dZ      dY 


i     j     k 

dx  dy  dz 
X    Y    Z 

dX 


dx) 


/dY    _  dX\ 
~  dill 


\dx       dy 

iCx  +  jCy  +  kC,. 

The  curl  of  a  vector  is  thus  a  vector  and  its  components  along 
the  axes  are  Cx,  Cv,  and  Cz. 

It  is  important  to  analyze  the  meaning  of  this  new  vector. 


dy 


c 


^Z, 


dz 


FIG.  139. 

Consider  a  small  rectangle  in  the  y-z  plane,  Fig.  139.  Let 
the  component  of  R  along  the  ^/-axis  be  Y  and  let  it  change  to 
FI,  as  we  move  along  the  z-axis  from  a  to  b. 


dz 


dz 


Similarly, 


257 


258  ELECTRICAL  ENGINEERING 

The  line  integral  around  the  rectangle  is  then : 
dL  =  Ydy  +  Zidz  -  Y^dy  -  Zdz 

idZ       dY\    ,    , 
=  I  -  -  —  —  )  dydz. 
\dy       dz  j 

Extending  this  to  all  three  planes,  we  get:  the  line  integral 
around  dS, 


-}dzdx  + 

dz        dx 


dy 

=  C  cos  adS,  where  a  is  the  angle  between  curl  C  and  the 
normal  to  the  surface  dS. 

The  z-component  of  the  curl  Cx  is  then  seen  to  be  the  limit 
of  the  ratio  between  the  line  integral  of  the  vector  around  a  small 
element  in  the  y-z  plane  and  the  area  of  the  element.  Since  it 
is  the  ^-component,  it  is,  of  course,  at  right  angle  to  the  surface, 
dydz. 

In  general, 

r.     i        r      AL        dL 
Curl  =  lim  -TO  =  -TO, 

AS       dS 

where  surface  dS  is  normal  to  the  vector  C. 

Stokes's  Theorem. — STOKES'S  theorem  states  that  the  line 
integral  of  a  vector  R  around  any  closed  contour  is  equal  to  the 
surface  integral  of  the  curl  of  the  vector  over  the  surface  or  cap 
enclosed  by  the  contour. 

The  theorem  holds  always  when  transforming  from  the  line 
integral  to  the  surface  integral,  but  applies  in  the  transformation 

from  the  surface  to  the  line  integral  only  when  -r—  +  — h 

-T—  =  0,  that  is,  only  when  the  curl  has  no  divergence. 

Depending  upon  the  system  of  notations  used,  it  is  written 
in  either  of  the  following  ways: 
In  vector  notation,  it  is: 

fR  •  dr  =  f  f  (V  X  R)  '  NdS, 

which  is  to  be  read:  The  line  integral  of  the  electric  field  in- 
tensity along  the  circuit  is  equal  to  the  surface  integral  of  the 
curl  of  the  vector  over  any  surface  (any  cap)  bounded  by  the 
circuit,  where  N  is  the  unit,  outward  drawn  normal  to  dS. 


THE  CURL  OF  A  VECTOR  259 

Obviously,  the  theorem  may  also  be  written: 

fRds  cos  (Rds)  =  f(Xdx  +  Ydy  +  Zdz) 
dZ       dY\  dX       dZ  BY       d 


where  I,  m  and  n  are  defined  below. 

The  theorem  can  best  be  proven  by  calculus  of  variations, 
but  may  be  understood  without  mathematics  by  the  following 
reasoning.  Refer  to  Fig.  140,  which  shows  the 
cap  divided  up  into  a  number  of  small  elements. 
It  is  evident  that  the  sum  of  the  line  integrals 
around  all  these  small  areas  resolves  itself  into  the 
line  integral  around  the  contour,  'since  all  lines, 
except  the  contour,  are  traced  in  two  equal  and  „ 
opposite  directions. 

Thus  if  dL  is  the  line  integral  around  one  of  the  small  areas, 
then 

2dL  =  fR  cos  (Rds)dS. 
But  it  has  been  shown,  that 

fdZ      dY\.  dX      dZ.  /BY 

dL  =     " 


\. 

-  dx)dzdx 


cos  adS,  where  a  is  the  angle  between  the  curl  C  and  the  normal 

to  the  surface  dS.  (2) 

.*.  dL  =  C  cos  adS' 

but  dydz  =  IdS,  where  I  =  cos  (Nx)'} 
dzdx  =  mdS,  where  m  =  cos  (Ny)  ; 
dxdy  =  ndSt  where  n  =  cos  (Nz); 

by  substituting  these  values  in   (2),   equation   (1)   is  proved. 


17 


CHAPTER  XXIV 
THE  EQUATION  OF  THE  ELECTROMOTIVE  FORCE 

It  has  been  shown  that  the  potential  difference  between  two 
points  in  an  electric  field  is  the  line  integral. 

V  =  f(Xdx  +  Ydy  +  Zdz)  =  fGds  (1) 

where  X,  Y  and  Z  are  the  components  of  the  field  intensities  or 
gradient  along  the  x,  y  and  z  axes  and  V  is  expressed  in  electro- 
static units. 

It  will  be  shown  later  that  the  conversion  factor  between  the 
electro-static  units  and  electromagnetic  units  of  potential  is  the 
velocity  of  light,  v  =  3  X  1010  cm.  per  sec. 

The  e.m.f.  in  the  electromagnetic  system  of  units  is  v  times 
that  in  the  electro-static  system  of  units.  Equation  (1)  should 
be  written: 

V  =  vf(Xdx  -f-  Ydy  +  Zdz)  in  electromagnetic  units     (2) 

Experiments  have  also  shown  that  the  e.m.f.  in  electromagnetic 
units  in  a  circuit  is  equal  and  opposite  to  the  product  of  the  turns 
enclosing  the  magnetic  flux  and  the  rate  of  change  of  the  flux. 

If  L,  M  and  N  are  the  components  along  the  x,  y  and  z  axes 
of  the  magnetic  field  intensity,  and  if  I,  m  and  n  are  the  direction 
cosines  of  the  normal  to  the  surface  dS,  and  if  /z  is  the  permea- 
bility then  the  flux  is: 

0  =  ffpQL  +  mM  +  nN)dS  =  ffpH  -  dS 
Then  the  e.m.f.  induced  per  turn  is: 

V  =     -  ~  =      -  ^  [ff»(lL  +  mM  +  nN)dS]          (3) 
combining  (2)  and  (3),  and  assuming  M  constant, 


But  from  STOKES'S  theorem,  we  can  write: 

f(Xdx  +  Ydy  +  Z*i)  =//[Kf  -  £)  +  »  (f  -  g 


260 


THE  ELECTROMOTIVE  FORCE  261 


Equating  (4)  and  (5),  we  get: 


dY\  /dX       dZ\          /dY       dX\-\ 

m  \~d~z  -  dx)  +  n  (to  "  a    J  = 


r 


~  v   dt 


(6) 


If  the  circuit  be  closed,  a  conduction  current  will  flow,  and  its 
magnitude  will  depend  upon  the  resistance. 

NOTE.  —  If  the  circuit  is  inductive,  this  applies  equally  well,  since  in  these 
equations  the  total  variation  in  flux  is  considered. 

Let  I,  with  components  u,  v  and  w  be  the  current  density 
along  the  x,  y  and  z  axes,  and  p  be  the  resistivity  of  the  ma- 
terial. Let  ds  with  components  dxt  dy  and  dz  be  an  element 
of  the  circuit,  and  Ax,  Ay  and  Az  be  the  projected  areas  of  an 
elemental  surface  dS,  then  the  resistance  along  the  z-axis  is 

-  dx  and  dV  =  —  (resistance  X  current)  =  --  T—  uAx  =  pudx, 

Ax  Ax 

but 

V  dV  •      Y 

X  =      ~~dx=pu>  -X  =  pu- 
Similarly,  Y  =  pv, 

and;  Z  =  pw. 

It  should  be  noted  that  X,  Y  and  Z  are  expressed  in  electro- 
static units.  Thus  by  transforming  the  relations  to  electro- 
magnetic units,  we  get: 

pu  =  vX] 
pv  =  zY', 
pw  =  vZ. 

The  Equations  of  the  Current.  —  Let  the  components  of  the 
current  density  along  the  three  axes  be  u,  v  and  w,  in  electromag- 
netic units.  Let  I,  m,  and  n  be  the  direction  cosines  of  the 
normal  to  surface  dS't  then  the  total  current  is: 

+  mo  +  nw)dS. 


262  ELECTRICAL  ENGINEERING 

It  was  shown  by  AMPERE  that  the  work  done  in  carrying 
unit  pole  around  an  element  carrying  current  i  was  4iri. 

The  work  done  is  J*  (Ldx  -f  Mdy  +  Ndz),  where,  as  usual, 
L,  M  and  N  are  the  components  of  the  magnetic  field  intensity. 

.*.  f(Ldx  +  Mdy  +  Ndz)  =  4x7  =  4*ff(lu  +  mv  +  nw)dS. 
But  by  STOKES'S  theorem, 

f(Ldx  +  Mdy  +  Ndz)  =  ff(lCx  +  mCy  +  nCJdS. 

+  mv  +  nw)dS. 


dN       dM 

=   Cx   =  -r  --    -5—  » 

a?/       a^ 
r      aL     aAr 

=  Cy  =  -  -  —  , 

and,  4       _  c    ._  5M  _  aL 

"   dz         dy 

Energy  of  the  Electric  Field.  —  Consider  a  small  cube-shaped 
volume  dxdydz,  Fig.  141,  in  the  electric  field,  and  let  the  po- 
tential difference  between  the  two  sides  dxdy  be  V. 


FIG.  141. 

The  capacity  of  the  field  enclosed  by  the  cube  has  been  shown 
to  be: 


M  J          M  J 

4ird          4irdz 

The  energy  stored  in  the  field   is  J^CF2,  and   the   potential 

dv 
V  is  Zdz,  where  Z  =  —  • 

_  Kte*y  ^  _  KZVxdydz 

4irdz  Sir 

KZ*    ,  KZ*     . 

-« —  dv,  or  the  energy  per  unit  volume  =   -^ — ,  when  only 

the  ^-component  of  the  field  is  considered. 


THE  ELECTROMOTIVE  FORCE  263 

If  the  components  of  the  electric  field  intensity  R,  are  X, 

TT 

Y  and  Z,  then,  the  total  energy  per  unit  volume  =  Wo  =  ^ 

O7T 

(X2  +  F2  +  Z2). 

Similarly,  it  is  proven  that  the  energy  stored  per  unit  volume 
in  the  magnetic  field  is: 

W0  =  ^  (L2  +  M*  +  TV2). 

Thus  the  total  energy  per  cubic  centimeter  in  space  occupied 
by  magnetic  and  electric  field  is: 

W0  =  ^  [M(L2  +  M2  +  JV2)  +  K(X*  +  F2  +  Z2).] 

There  appears  to  be  no  limit  to  the  possible  intensities  of  the 
magnetic  field,  but  for  the  electric  field  in  air  at  atmospheric 
pressure,  experiments  indicate  a  maximum  possible  gradient,  or 
field  intensity  of  30,000  volts  per  cm.,  or  100  electro-static 
units  of  potential  per  cm. 

Thus  in  the  electric  field  the  maximum  amount  of  energy  at 
normal  pressure  is: 

1002 
Wmax.  =  -~ — -  =  400  ergs  per  cu.   cm.   or  0.00004  joules  per 

cu.  cm. 

Maxwell's  Displacement  Current. — MAXWELL  assumes  that 
when  a  potential  difference  exists  in  any  part  of  a  dielectric,  an 
electric  displacement,  or  a  displacement  of  electricity  has 
taken  place  along  the  lines  of  electric  intensity  (force).  The 
greater  the  displacement,  the  greater  the  difference  in  potential. 

The  displacement,  however,  is  resisted  by  the  electric  elasticity 
of  the  medium,  which,  for  the  lack  of  a  more  satisfactory  analogy, 
can  be  thought  of  as  being  in  a  way  similar  to  that  existing  in 
an  elastic  body,  against  which  a  particle  is  pressed. 

For  a  given  potential  difference,  the  displacement  is  greater 
the  greater  the  specific  inductive  capacity;  for  example,  if  the 
dielectric  be  glass,  the  displacement  may  be  five  to  six  times  as 
great  as  would  be  true  with  air  or  vacuum. 

A  metal  may  be  considered  to  have  zero  capacity,  in  other 
words,  energy  can  not  be  stored  into  it,  but  electricity  would 
continue  to  pass  through  it  as  long  as  a  potential  difference 
existed. 

Dielectrics,  on  the  other  hand,  would  permit  electricity  to 
flow  up  only  to  a  certain  distance,  and  the  flow  ceases  when  the 


264  ELECTRICAL  ENGINEERING 

force  causing  the  electricity  to  flow  is  exactly  equal  to  the 
opposing  force  due  to  the  elasticity  of  the  dielectric. 

The  displacement  of  electricity  is  in  the  direction  of  the  lines 
of  electric  force;  since  the  displacement  has  magnitude  as  well 
as  direction,  it  is  a  vector  quantity. 

According  to  MAXWELL'S  theory  an  electric  current  is  a  time 
rate  of  change  of  the  displacement  of  electricity. 

The  charge  on  a  body  is  a  measure  of  the  displaced  electricity. 
Indeed,  MAXWELL  states  that  a  charge  Q  on  a  body  causes  a 
displacement  of  Q  units  of  electricity  out  from  the  body,  and  he 
has  defined  the  displacement  D  as  the  charge  per  unit  area.  It 
is  then  numerically  equal  to  a1,  the  charge  per  unit  area,  but 
while  or  is  a  scalar  quantity,  D  is  a  vector. 

D  can  be  expressed  as  a  function  of  the  intensity  R  and  the 
specific  capacity  K. 

T       •    ^     •  j.       •*.       i?   xi_     />  i  j    •      flux          ^         4?rC      T 
In  air  the  intensity  of   the  field   is  -  -  =  —j—     In 

area       area         A 

other  dielectric  of  specific  capacity  K, 

P        1          47TQ  ARK 

R  =  K         ~T         "Q~-    ^T 

Q       ARK       RK 

The  surface  charge  =    T  =  -A 
A        4: 


Thus  the  displacement  D  is  also, 


The  displacement  of  electricity  is  in  the  direction  of  the  field. 
Thus  if  /,  g  and  h  are  the  components  of  the  displacement,  and 
X,  Y  and  Z  are  the  components  of  the  electric  field  intensity, 
then, 

KX 


g 


47T    ' 

^^  In    these    equations,    the    units 

are  in  the  electro-static  system, 
and, 

,       KZ 
h  =  -T—. 
4?r 

The  amount  of  electricity  displaced  is  the  product  of  current 
and  time,  or  considering  current  per  square  centimeter  or  current 
density,  the  displacement  is  the  product  of  current  density  and 
time. 


THE  ELECTROMOTIVE  FORCE 


265 


Let  Ud,  vd,  and  wd  be  the  components  of  the  current  density, 
then: 

ud  dt  =  df, 
Vd  dt  =  dg, 


and, 


dt  =  dh. 


It  has  been  shown  that  the   conduction  current  density  in 
electro-static  units  was: 
X 


and, 


Z 

w  =  — , 


where  p  is  the  specific  resistance. 


Thus  the  total  current  density  along  the  x-axis  is: 
similarly, 


=  X   ,df  =  X      JtdX 
p         dt         p         4ir   dt  ' 


Y   ,    dg  Y       KdY 

v  +  vd  = h  -7.-  = h  T-  -7:7 

p         at  p        4?r   at 

Z        dfc  Z        X  dZ 


Thus,  applying  AMPERE'S  relation,  that  in  electromagnetic 
units  the  curl  of  the  magnetic  field  intensity  is  4?r  times  the 
current  density,  we  get  : 


47T    , 

-  (w 


similarly, 
and, 


1/47TZ 
9\    p 


dX\    _  lr47T 
K  dt)   :=  vlp 


- 
dt 


dM 


dz        dx 
dM      dL 


(16) 


L  p  dtJ  dx         dy 

where  v  at  present  is  the  unknown  ratio  between  the  units. 
The  corresponding  equations  for  the  e.m.f.  were  shown  to  be 

•jT  *\  rr  *\  ~\7 

fj,  (JLJ  O  £j  O  I 

~  v  dt          dz     ~  dx 


266  ELECTRICAL  ENGINEERING 

v  dt     =f  ~?x  ^ 

_  p  cW    _  dY  _  dX 
~  v  dt      =  dx  "  dy 

By  combining  equations  (1)  and  (2),  it  is  possible  to  arrive  at 
equations  of  the  electric  and  magnetic  field  intensities  in  any 
medium — conductor  or  non-conductor. 
Differentiate  (la)  with  respect  to  t, 

7v  7"  ~dt  +  K  ~W      =  dydt  ~  ~dzdt 
Differentiate  (2c)  with  respect  to  y, 

IJL  d2N        d2Y        d2X 
*  '  ~~  ~v  ~didy  =  dxdy  ~  'dy2 
Differentiate  (26)  with  respect  to  z, 

fi  d2M       d2X        d2Z 
"  ~v  Htdz  ==  'dz2   ~  dxdz  ^ 

«2  ~y 

Substitute  (4)  and  (5)  in  (3),  and  add  and  subtract  — ^  = 


T—  (  -^—  }  ,  the  following  equation  results: 

47TM  dX  32X          rd*X    .^X.^X        d_ 

p     dt   '    AM  dt2    ''    V  Idx2  "  dy2  "  dz2   '    dx 


which  is  the  most  general  equation. 

If  there  is  no  divergence,  that  is  if  we  are  interested  in 
medium  having  no  charges,  then  the  equation  becomes: 


It  is  readily  seen  that  exactly  similar  equations  not  only 
result  for  the  Y  and  Z  components  of  the  electric  field  intensity, 
but  also  for  the  components  of  the  magnetic  field  intensity, 
L,  M  and  N. 

Special  Cases.  —  (a)  In  a  dielectric,  p  =  »  ,  thus  the  equations 
become  : 


(8) 


THE  ELECTROMOTIVE  FORCE  267 

or  in  general, 

-Qp    =  a2V2?7,  where  a2  =  ~,  and   U  stands  for  either  X,   Y, 
Z,  L,  M  or  N. 

This  is  the  well-known  equation  of  the  propagation  of  any 
disturbance  at  finite  speed. 

The  velocity  of  the  propagation  is  a  =  -   —.      In   air.   k  =  1 


and  M  =  1,  thus  the  velocity  of  propagation  of  the  electric 
and  magnetic  field  is  v. 

This  value  has  been  measured  and  found  to  be  that  of  light, 
thus  the  conversion  factor  is  the  velocity  of  light.  Thus  v  = 
3  X  1010.  ' 

This  important  fact  was  deduced  by  MAXWELL  in  1865. 

(6)  In  a  conductor,  the  specific  inductive  capacity  may  be 
assumed  as  zero,  thus  we  get: 


or, 

d*U   ,   d2U      47r  dU   . 
d^+^=^-arin  rectangular 

coordinates,  and, 


,  ,    1  dU  . 

^   a0T  +  -^  +  r  -^T  ==  >  ln  cylindrical 

coordinates. 

; 

Assuming,  as  an  application,  that  it  is  desired  to  determine 
the  current  distribution  at  any  time  in  a  cylindrical  conductor 
at  any  distance  from  the  origin  and  any  distance  from  the 
center  of  the  conductor.  If  the  practical  system  of  units  is 
used,  v2  =  1;  and  on  account  of  circular  symmetry,  the  term 

involving  —  disappears.     Thus  the  equation  becomes: 

dzi        dzi       1    di       4?r  di  ,     , 

a72  +  a^2  +  r  dr  =  7  Hi 

Distribution  of  current  in  a  cylindrical  conductor:  If  it  is  of 
interest  to  find  the  distribution  along  a  radius  only,  the  equation 
becomes  : 

dH       1    di  _  47r  di  ,     . 

dr2  +  r  dr  ~  p     dt 


268  ELECTRICAL  ENGINEERING 

It  will  be  of  interest  to  verify  this  equation  directly.  It  has 
been  shown  that  the  work  done  in  ergs,  in  taking  unit  pole  once 
around  a  conductor  carrying  current  7  is  4?r7,  where  7  is  the 
current  enclosed  in  the  path. 

Consider,  for  the  sake  of  simplicity,  a  cylindrical  conductor, 
Fig.  142.  Let  the  instantaneous  values  of  the  current  density 

at  distant  r  from  the  center  be  i,  and  that  at  r  +  dr  be  i  +  —  dr. 


FIG.  142. 
Let  the  magnetic  field  intensity  at  distant  r  be  H]  and  at 

riff 

distant  r  +  dr,  be  HI  =  H  -f  —  dr. 

The  work  done  on  unit  pole  in  going  from  a  to  b  is: 

#i(r  +  dr)6  =   (H  +  ^  dr)   (r  +  dr)8  = 

(•\TT  v 

Hr  +  #dr  +  r  -~-  dr\  ,  neglecting   the  term   which   in- 

volves (dr)2. 

The  work  done  in  going  from  b  to  c,  or  from  d  to  a,  is  zero, 
because  we  travel  on  an  equipotential  surface. 

The  work  done  in  going  from  c  to  d  is  —Hrd. 


:.  W  =  e( 


Hdr  +  rdr    =  edr 


And  by  definition  given  above, 

W  =  4irir0dr,  neglecting  the  term  which  involves  (dr)2. 

H      dH 


/1x 
(1) 


THE  ELECTROMOTIVE  FORCE  269 

The  ohmic  drop  in  voltage  along  1  cm.  of  the  conductor  at 
the  outer  edge  of  the  segment,  that  is,  at  r  +  dr  from  the  center, 
perpendicular  to  the  paper,  is: 

(i  +  —  dr)  p,  where  p  is  the  specific  resistance. 

The  drop  along  the  inner  edge  is  ip;  thus  the  difference  in  the 
e.m.f.  at  the  two  edges  is: 

de  = 

This  must  be  then  the  e.m.f.  which  is  consumed  by  the  self- 
induction  due  to  the  flux  in  the  element. 

The  flux  in  the  element  is  0  =  pH(dr  X  1  cm.)  ==  Hdr      (3) 

AA.  AH 

(4) 
(5) 


From  (2)  and  (4), 

di  dH  di  _     dH 

Differentiating  (1)  with  respect  to  t, 

di  __  1  dJH       d*H 
'         dt      r     dt  +  drdt 

Differentiating  (5)  with  respect  to  r, 


Substitute  (7)  in  (6), 

di       1  dH       p  dH 

.*.  4?r  —  =  _-  —  +  -  —z 

Substitute  the  value  of  ---  from  (5)  in  (8), 

di  _  1    p  di  p  dH 

dt       r  n  dr  IJL  dr2 
or, 

M       dH   .  1    di 


_ 
p     dt  ~  ar2  ""  r  dr 

in  electromagnetic  system  of  units. 

Equation  (9)  is  very  important  in  connection  with  problems 
of  heat  as  well  as  electricity,  it  has  been  studied  by  great  mathe- 
maticians, notably,  MAXWELL  and  LORD  RAYLEIGH. 

It  is  to  be  noted  that  the  right-hand  member  of  equation  (9) 
is  LAPLACE'S  equation  transformed  to  cylindrical  coordinates, 


270 


ELECTRICAL  ENGINEERING 


when  the  cylinder  has  circular  symmetry.     Thus  we  could  have 
written  : 


Special  Case.  —  Flat  bar:  Referring  to  Fig.  69,  in  the  case  of 
flat  bar,  r  approaches  infinity,  and  (9a)  becomes: 


Si 


FIG.  143. 

The  distribution  of  flux  in  a  cylindrical  conductor  surrounded 
by  an  energized  solenoid  is  determined  in  a  similar  way.  Fig. 
143  shows  the  path  of  the  current  and  flux.  The  dots  represent 
the  current,  and  the  lines  around  the  current,  the  flux. 

The  result  for  a  cylinder  is  identical  with  equation  (9),  if 
H  is  substituted  for  i. 

Similarly,  for  a  flat  bar  equation  (10)  is  applicable  with  the 
same  substitutions. 


CHAPTER  XXV 

MATHEMATICAL  SOLUTION    OF  EQUATION   11,    PAGE 

267,  DEALING   WITH   ALTERNATING   CURRENT 

DISTRIBUTION    IN     CIRCULAR    CYLIN- 

DRICAL CONDUCTOR 

The  general  equation  is  as  has  been  shown: 

dH       1   di      47TM  di 
dr2  +  r  dr  ~     p     dt 

Since  we  are  dealing  with  sine  waves,  let: 

i  =  i\  cos  cot  +  z*2  sin  cot  (2) 

where  ii  and  iZj  the  current  densities,  are  functions  of  r  but  not  of 
t.     Substitute  first,  i  =  i\  cos  cot, 

di  dii 

-  =cosa^-> 


-2 

and>  fU  -a*  sin  erf. 

at 

dH\  1  dll  —    4:TTfJL    . 

.'.  cos  ut  T-^-  H  —  cos  (^t  T—  =—  -  iico  sin  cot  (3) 

Similarly,  for  i  =  i%  sin  co£, 

J    a2l2     ,     1  dll          +47TM     . 

sin  ut  T—  2-  +  -  sm  o)t  —  =  -  izoj  cos  cot  (4) 

Adding  (3)  and  (4), 


•'•  COS  «M  ^  +  -    fr-     —  H| 


and  d2iz       1   5i2 

W+r   dr= 
271 


272 


ELECTRICAL  ENGINEERING 


+ 


62r2 


Assume : 

i\  ==  a<)  ~\~  air 

and 

Then: 

v-1  =  01  +  2a2r  +  3a3r2 


-=-£-  =  2a2  +  6a3r 
Let 


•  •  + 
n(n  — 


bnrn 


.n-l   + 


•HI' 


(8) 
(9) 

(10) 
(11) 

(12) 

(13) 
(14) 

(15) 


'.  ai  =  0;4a2  =  m260;9'a3  =  7^26  1;  in  general,  n2an  =  m26n_2  (16) 

By  similar  substitutions  in  (14),  we  have: 
1  =  Q;  462  =  —  m2a0;  963  =  —  m2aij  in  general, 


(6)  and  (7)  can  be  written: 


dr 


and, 


dr 


dr 


Substituting  (9),  (10)  and  (11)  in  (13), 

2a2r  +  6a3r2  +  •  •  •  +  n(n  —  l)anrn~l  - 
+  a!  +  2a2r  +  3a3r2  +  •  •  •  +  nanrn~l 


6n  =  -  m2an_2,  or  (n  -  2)2  &n_2  =  -  m2  an_4 
Combining  the  last  equations  in  (16)  and  (17), 

m4 


an  =  — 


From  (17), 


n2(n  -  2) 


(17) 


(18) 


(19) 


Since  ai  =  0,  and  61  =  0,  from  (18)  and  (19)  all  the  a's  and  6's 
with  odd  indices  separately  equal  to  zero.  And  those  with  even 
indices  are  as  follows: 


SOLUTION  OF  ALTERNATING  CURRENT        273 


ac  =  a0 

m4 

w4 
as  =  —  fi-FToS  a 

and  so  forth 


m 


42 . 62  «*• 

m4  m8 

•  82  .^02  «6  =    +  42"7g2Tg2~ 


and  so  forth 


(19)] 


60    =+  7^02  [See  (16)] 


a0 


m 


10 


22-42-62-82 


7o2«o 


and  so  forth 


m2 
82 
m2 


m6 


42-62-82-(10)2-(12)2 


and  so  forth 


Therefore,  i\  =  a0  (l  ~  02742  +  "02.  42.^2.02  ~  '  *  ' ) 


H 9 


, 

"t" 


and 


52         22-42-62       22-42-62-82-(10)2 
~2"2~   ~  22-42-62  +  ***]  + 


m4  r4 


)    (21) 


LORD  KELVIN  has  denoted  the  first  series  in  (20)  by  ber  (mr) 
and  the  second  in  (20)  by  bei  (mr),  thus: 


oer(x) 


22-42-62-82 


,  and 


/v.2  ^.6  /rlO 

,.XN         *>  ^  |  •t' 

=  2*~  22T42T62  "*"  22"T42  •  62  •  82  •  (lO)2 

And 

4 

ii  =  a0  for  (mr)  +  — 2  a2  6et  (mr),  and, 

4 

t'2  =  —  ao  6ei  (mr)  +  — 5  a2  6er  (mr) 


(22) 


(23) 


These  functions,  6er  and  6^',  have  been  worked  out  and  appear 
frequently  in  books  on  mathematical  physics. 


274 


ELECTRICAL  ENGINEERING 


Therefore, 

i  =     a0  her  (mr)  -\ -2  a2  bei  (mr)     cos  ut  + 

\—£ az  ber  (mr)  —  a0  bei  (mr)  I  sin  cot       (24) 

The  constants  a0  and  a2  are  determined  from  the  fact  that  the 
extreme  outside  layer  is  not  surrounded  by  any  flux.  (We 
consider  only  the  flux  in  the  wire  in  this  calculation.)  Thus  the 
sine  term  is  zero  at  all  values  of  t. 

Let  Jo  =  maximum  value  of  the  current  density  at  the  surface, 
then, 


and 


70  =  a0  ber  (mR)  -\ ^  bei  (mR), 


2 

0  =  — ^  her  (mR)  —  a0  bei  (mR) 


(25) 


Equations  (25)  are  readily  solved  and  give: 
4a2  _  JQ  bei  (mR) 

m?   ~~~  ber2  (mR)  -f  bei2  (mR)' 

IP  ber  (mR) 
ber2  (mR)  +  bei2  (mR) 

{[ber  (mR)  ber  (n 


and, 


bei 


(26) 


bei 


ber2  (mR)  +  bei2  (mR) 
(mr)]  cos  ut  +  [bei  (mR)  ber  (mr)  —  ber  (mR)  bei  (mr)]  sin  co£}  (27) 
Thus  the  square  of  the  effective  current  density 


I  r)P7*^     I  »?  rr   I     r>^7*^     ( 'YyiT'i      — 

-j~\\ TO  [t/C'/         \^ffvL\j)     Ut/l         \ifvl  ) 

ber2  (mR)  bei2  (mr)  +  bei2  (mR)  bei2  (mr)  +  bei2  (mR)  ber2  (mr)] 
7°2  [ber2  (mr)  +  bei2  (mr)]. 


2[ber2  (mR)  +  bei2  (mR)] 


V2 


(mR)  +  bei2  (mR) 


(28) 


At  the  center  of  the  conductor,  r  =  0, 

.'.  ber  (mr)  =  1, 

bei  (mr)  =  0. 
1 


•  leff.    = 


_ 

A/2  Vber2  (mR)  +  bei2  (mR) 


at  r  =  0. 


SOLUTION  OF  ALTERNATING  CURRENT        275 

With  very  low  frequency,  the  current  density  approaches  the 
direct  current  case  where  it  is  normal  and  is: 


thus  the  ratio  of  the  alternating  current  density  at  the  center. 
to  that  of  the  direct  current  is: 

_  __  1  __ 

\/ber2  (mR)  +  bei2  (mR) 

For  copper, 

fj.  =  1,  =  and  p  =  1600, 


If  the  radius  is  1  cm.,  and  the  frequency  is  60, 

mR  =  1.72, 

[ber2  (mR)  +  bei2  (mR)]~*  =  0.87. 

.'.  the  current  density  at  the  center  is  87  per  cent,  of  that  at 
the  surface,  and  also  87  per  cent,  of  what  it  would  be  with 
direct  current. 

If  the  conductor  had  a  diameter  of  50  cm.,  the  current  density 
at  the  center  would  only  be  25  per  cent,  of  that  at  the  surface. 

Actual  watts  consumed  in  heat  are : 

'72 

[ber2  (mr)  +  bei2  (mr)]d(r2) 

-(29) 


C 
2  I 

Jo 


2[6er2  (mR)  +  bei2  (mR)] 
W  =  (ohmic  resistance)  •  (total  eff.  current)2  (30) 

Ohmic  resistance  =  -™  (31) 

Total  current  =  irK 


i2irrdr  = 


ber2  (mR)  +  bei2  (mR) 
ber  (mR)  \    ber  (mr)d(r2)  +  bei  (mR)  \    bei  (mr)d(r2)   • 

r        CR  CR         i 

Cosut-\-\bei(mR)  I  ber(mr)d(r2)  —  ber(mr)  I  bei(mr)d(r2)   sincoZ  • 
L  Jo  Jo  J 

272 

.'.  (total  eff.  current)2  =  or,     ,  . — PN    ,°  ,    .9  . — ^r\- 

2[ber2  (mR)  +  bet2  (mR)] 

\  CR  I2     \  CR  121 

I    ber  (mr)d(r2)      +      j  bei  (mr)d(r2)  (32) 

U  Uo 


276 


ELECTRICAL  ENGINEERING 


7rp/o2j|Jo    6er(mr)d(r2)J2+|Jo 

bei  (mr)d(r2)      \ 

2R2[ber2  (mR)  +  bei2 

(mR)] 

The  coefficient  of  skin  effect  = 

wact. 

'  CR            2      CR 

Jo                                Jo 

2  (mr)d(r2)  \R2 
J 

W 

jf  ber  (mr)d(r2)\2+  \  i 

I2 

bei  (mr)d(r2) 

(33) 


(34) 


(mr)  =  1  -  312  +  25,500 


210,100©% 


(mr)  +  bei1  (mr)  ==  1  +  313  +  3900 


I     [6er2 


/mr\  12 
21,700  Q      + 


(mr) 


3100 


(35) 


CR 

\     ber 

° 


J 


'  (mr)d(r2)  = 


I  I 


SOLUTION  OF  ALTERNATING  CURRENT         277 


Substituting  (35)  (36)  and  (37)  in  (34), 


K  = 


10 


(38) 


The  following  tables  give  the  coefficient  of  skin  effect  at 
various  values  of  mil  and  the  values  of  m  for  copper,  aluminium 
and  iron. 


mR 

K 

mR 

K 

mR 

K 

0 

1 

3.0 

1.32 

6.0 

2.39 

0.05 

1.0001 

3.5 

1.49 

8.0 

3.10 

1.0 

1.005 

4.0 

1.68 

10.0 

3.79 

1.5 

1.026 

4.5 

1.86 

15.0 

5.57 

2.0 

1.08 

5.0 

2.04 

20.0 

7.32 

2.5 

1.17 

5.5 

2.22 

Material 

M 

p  in.  e.m.u. 

m 

Copper.  . 

1 

1700  at  20  °C. 

0  216  A/7 

Aluminium 

1 

3,000 

0  162  A/7 

Iron  

300  to  1200 

10,000 

0  .  09    A/M? 

The  value  of  /x  for  iron  is  usually  taken  as  300,  but  experi- 
ments on  iron  wires  used  as  transmission  lines  seem  to  give  values 
of  M  as  high  as  1200. 

LORD  RAYLEIGH  has  shown  that  when  the  penetration  is  so 
slight  that  the  above  table  can  not  be  used  a  close  approximation 
of  the  " effective  thickness"  in  centimeters  of  the  surface  layer 
which  causes  the  current  is: 

0     ==  7  


where  K  is  the  specific  conductivity. 

6.6 


This  formula  becomes  6  = 


—7=     for  copper  approximately. 

8  8 

— ^=    for  aluminium  approximately. 


16 

Vrf 


for  steel  approximately. 


CHAPTER  XXVI 
ELECTROMAGNETIC  RADIATION 

Introduction. — The  laws  governing  electromagnetic  radiation 
were  stated  by  MAXWELL  fifty  years  ago.  The  experimental 
verification  was  presented  twenty  years  later  by  HERTZ  in  a 
series  of  most  extraordinary  papers,  which  were  later  published 
in  book  form.  The  practical  application  was  made  by  MARCONI. 

An  extensive  literature  is  now  available,  notably  FLEMING'S 
"The  Principles  of  Electric  Wave  Telegraphy  and  Telephony," 
and  ZENNECK'S  "Wireless  Telegraphy." 

In  writing  this  chapter  the  author  has  drawn  extensively  upon 
the  information  which  is  given  in  these  books.  Since  it  is  likely 
that  students  who  have  not  read  what  preceded  this  chapter 
will  want  to  understand  the  principles  of  wireless  transmission 
it  has  seemed  wise  to  built  up  the  theory  from  the  fundamental 
laws  even  though  this  procedure  necessarily  involves  some 
repetition  of  what  has  been  given  in  previous  chapters. 

Fundamental  Conceptions. — Surrounding  any  body  charged 
with  electricity  is  an  electric  field.  The  intensity  of  the  field 
usually  varies  from  point  to  point,  but,  at  any  point  it  is  propor- 
tional to  the  charge,  that  is,  the  amount  of  electricity  on  the 
charged  body. 

To  charge  a  body  we  connect  it  to  a  source  of  potential  when  a 
current  momentarily  flows  from  the  source  to  the  body,  the  cur- 
rent stopping  when  the  potential  of  the  body  is  the  same  as  the 
potential  of  the  source. 

If  i  is  the  current  flowing  during  an  interval  of  time  dt  then 
the  resulting  charge  on  the  body  is  dq  =  idt,  or, 

i  -  ^ 
1  ~  dt 

For  reasons  that  will  appear  later,  it  has  been  assumed  that 
the  outward  field  of  flux  from  a  body  charged  with  Q  units  of 
electricity  is 

\l/  =  4wQ  lines  of  electric  force. 

278 


ELECTROMAGNETIC  RADIATION  279 

If  the  lines  are  uniformly  distributed  over  a  closed  envelope  of 
area  A  sq.  cm.,  then  the  density  of  the  electric  field  is 


By  the  introduction  of  the  constant  4ir  in  the  flux  formula  this 
density  becomes  in  space  the  same  as  the  force  in  dynes  per  unit 
charge  which  is  numerically  the  same  as  the  intensity  R  of  the 
electric  field  at  the  particular  point  considered.  This  is  easily 
seen  from  COULOMB'S  law,  which  states  that  the  repulsive  force 
between  two  charges  Q  and  Qi  is 

f_QQi 

J    -     Kr2 

where  r  is  the  distance  between  them  and  Vi  =  1. 

In  the  ideal  case  the  charge  is  confined  to  a  point  and  the  flux 
is  distributed  uniformly  in  every  direction. 

.  R  =        _±_        =^Q  =  Q 

area  of  sphere       4?rr2       r2 

where  r  is  the  distance  from  the  point  to  the  point  charge. 
or,  Q  =  Rr* 

/.  /  =  j3  r«Q!  =  RQ,. 

If,  therefore,  Qi  ==!,/=  R. 

The  potential  difference  between  two  points  in  an  electric 
field  is  by  definition  numerically  the  same  as  the  work  done  in 
moving  unit  charge  from  one  point  to  the  other. 

Thus,  if  X  represent  the  intensity  of  the  electric  field  in  a  cer- 
tain direction,  say  a  direction  parallel  to  the  x-axis  in  a  rectangu- 
lar coordinate  system,  then  the  potential  difference  across  a 
short  element  dx  is  dV  =  Xdx  =  force  on  unit  charge  at  dis- 
tance x,  or, 

Y_dV 
=  dx' 

Similarly  v       dV 

=  dy 
and  „  _  dV 

£J      —          7       * 

dz 

Y  and  Z  being,  respectively,  the  electric  intensities  along,  or 
parallel  to,  the  s&-a&d-y  axes. 

*     * 


280 


ELECTRICAL  ENGINEERING 


If  we  desire  to  find  the  potential  difference  between  the  ends 
of  a  wire  bent  in  a  small  rectangle  in  the  x-y  planes  and  the  inten- 
sities along  the  x  and  y  axes  are  X  and  X\,  Y  and  FI,  then  refer- 
ring to  Fig.  144, 

dV  =  Xdx  +  Yidy  -  X,dx  -  Ydy  (3) 


dy 


dx 

FIG.  144. 


For 


y  =  0 
y  =  dy 


X  =  X 
X  =  X, 


The  rate  of  change  of  X  as  we  travel  along  the  ?/-axis  is 

thus  the  total  change  in  distance  dy  is : 

dX, 

-r—  dy 
dy    y 


Similarly, 


Substituting  (4)  in  (3)  we  get 


dX 


(4) 


It  is  one  of  the  properties  of  the  electric  field  alone  when  free 
from  charges  that  the  above  potential  difference  is  zero  in  a 
closed  circuit. 

If,  however,  an  e.m.f.  is  induced  in  the  rectangular  circuit  by 
change  of  flux  treading  through  the  circuit,  then  we  get: 


dt 


dN  1 1 

-^dxdy 


ELECTROMAGNETIC  RADIATION 


281 


where  N  is  the  density  of  the  magnetic  field  perpendicular  to  the 
plane  of  the  electric  circuit. 

By  a  similar  reasoning  we  get  then  the  following  three  impor- 
tant equations: 

dX 
dy  = 


ar 

dx 
dX 

dz 


dy 


_ 

dx 
dY_ 

dz 


dN 
dt 
dM 
dt 
dL 
dt 


(5) 


where  X,  Y,  Z,  L,  M,  N,  are  respectively  the  electric  intensities 
and  magnetic  intensities  in  the  same  system  of  units  parallel  to 
the  x,  y  and  z  axes. 

Note  that  in  air  the  densities  are 
'the  same  as  the  intensities. 

The  next  consideration  is  in  re- 
lation to  the  magnetic  effect  of  a 
current. 

Let  A,  Fig.  145,  represent  the 
end  view  of  a  wire  A  carrying  a 
certain  current  dl,  perpendicular 
to  the  plane  of  the  paper.  Let  the 
curved  line  be  in  the  plane  of  the  paper. 

The  magnetic  field  intensity  at  P  is  then  //  and  this  is  defined 
similarly  to  R  as  numerically  the  same  as  the  force  on  unit  pole. 
Let,  therefore,  a  pole  of  unit  strength  be  carried  along  the  curved 
path,  Fig.  145.  The  work  done  per  unit  pole  in  completing  the 
journey  once  is  evidently 

W  =  fll  cos  6ds  =    (  -   -  cos  0ds 
J     r  • 


FIG.  145. 


but 


rda  rda 

-3 —  =.  cos  6    ,  .  as  = 


cos  6 


ra-2* 

:.  w  : 

Jtx  =  0 


2dlda  = 


The  work  is  independent  of  the  position  of  the  current  element 
and  the  path.  Thus  if  there  are  a  number  of  filament  currents 
inside  the  path, 

I  =  Zdl 
W  =  47rJ  (6) 


282  ELECTRICAL  ENGINEERING 

This  quantity  is  by  physicists  called 
^  the  magnetomotive  force,  around  the 

circuit,  whereas,  engineers  would  call 
I    Ll         it  47r  X  m.m.f  . 

d  Consider  now  a  small  rectangular 

surface,  Fig.  146,  in  the  x-y  plane  of  a 
magnetic  field,  and  let  L  and  LI,  M 
*     and   MI   be   the   components  of  the 
FIG.  146.  magnetic  field  intensities,  along  the  x 

and  y  axes  respectively. 

Then  the  line  integral,  or  work  on  unit  pole  around  the  element 
is 

Ldx  -\-  Midy  —  Lidx  —  Mdy. 

The  rate  of  change  of  L  as  we  travel  along  the  y-axis  is  ^-» 
thus  the  total  change  is  —  dy,  thus 

Ll=L  +    -  dy. 
Similarly 


dM  dL  idM      dL\ 

-  -  dxdy  =  (d-  -  ~)  dxdy. 


From  (6)  it  is  seen  that 


dy 

where  Iz  is  the  total  current  flowing  through  the  rectangle  per- 
pendicular to  dxdy. 

Depending  upon  the  medium,  this  current  may  be  the  ordinary 
conduction  current  such  as  flows  in  a  wire  or  the  charging  current 
which  is  incident  to  a  change  in  the  electric  field,  or  indeed,  the 
sum  of  the  two  currents. 

In  this  analysis  it  will  be  assumed  that  the  air  surrounding  the 
oscillator  is  free  from  ionization,  so  that  its  resistance  is  infinite; 
thus  the  only  currents  considered  are  the  "  displacement,  or 
charging  currents." 

MAXWELL  assumed  that  surrounding  a  charged  body  is  an 
electric  field,  the  strength  of  which  is  proportional  to  the  charge, 
and  that  the  intensity  of  the  field  is  a  measure  of  what  he  calls 


ELECTROMAGNETIC  RADIATION  283 

displaced  electricity.  The  displacement  of  electricity  is  in  the 
direction  of  the  field  intensity,  and  is  thus  a  directed  quantity. 
Numerically  a  charge  Q  displaces  Q  units  of  electricity  outward 
from  the  body.  Since  dQ  =  idt,  it  follows  that  the  displacement 
current  or,  as  engineers  say,  the  charging  current  is  proportional 
to  the  time  rate  of  change  of  the  electric  field  intensity. 

Or, 

dR 


where  R  is  the  intensity  and  a  a  constant  to  be  determined. 

MAXWELL  worked  out  his  theory  on  the  basis  that  the  dis- 
placement is  numerically  the  same  as  the  charge  per  unit  area. 

Thus 


area 


But  the  outward  normal  flux  from  a  charge  Q  is  \j/  =  4-n-Q ;  thus 
the  intensity  of  the  field  is 

R 


area       area 

r> 

.'.  R  =  4:ird  or  d  =  7-  in  air. 

•  i  =  —  ~ 

4?r  dt 

where  i  is  the  current  per  unit  area  or  current  density. 

If,  therefore,  u,  v  and  w  are  the  components  of  the  displace- 
ment current  densities  along  the  x}  y  and  z  axes  and  X}  Y  and  Z, 
the  components  of  the  electric  intensities  then : 

1    dX  1    dY  1    dZ 

u  =  -7—  — »  w  =  -: —  and  w  =  -A TT  (8) 

4-7T    d£  4;r    ^^  4.w    dt 

everythir^g  being  given  in  electro-static  units. 

From  (7)  and  (8)  it  is  evident  that  one  can  write 

/dM       dL\    .  dZ  _ 

( — ^r—  I  dxdy  =  4:irwdxdy  =  —  dxdy. 

\  dx        dy]  dt 

or, 

dM  _  dL       dZ 

dx   '    dy  ==   dt' 


284 


ELECTRICAL  ENGINEERING 


dN 

dM 

dX 

dy 
dL 

dz 
dN 

dt 
dY 

z 

dx 

dt 

dM 

dL 

dZ 

dx 

dy 

dt  1 

By    a    similar    reasoning   are    obtained   the    following    three 
relations. 


(9) 


everything  being  given  in  the  same  system  of  units. 

The  simplest  form  of  oscillator,  or  rather  that  form  which 
lends  itself  to  the  simplest  mathematical  treatment,  is  that  used 
by  HERTZ. 

The  oscillator  consists  of  two  large  spheres  separated  by  a 
considerable  distance  and  connected  by  wires  through  the  spark 
gap  to  the  source  of  energy  as  shown  in  Fig.  147. 


o 


Magnetic  Field 


FIG.  147. 


O 


It  will  be  assumed  that  the  electric  field  is  due  to  the  spheres 
alone,  and  the  magnetic  field  to  the  linear  conductor. 

It  will  be  assumed  that  the  axis  of  the  oscillator  is  the  2-axis. 
Thus  the  magnetic  field  which  is  in  the  form  of  rings  around  the 
conductor  has,  in  the  x-y  plane,  no  component  in  the  direction 
Z  and  therefore  no  e.m.f.  can  be  induced  in  the  x-y  plane. 
However,  e.m.fs.  will  be  induced  in  the  direction  of  the  Z-axis. 

Whatever  the  potential  distribution  in  the  x-y  plane  it 
must  thus  be  due  to  the  charges  on  the  spheres  alone,  that  is, 
due  to  the  electric  field  alone. 

The  distribution  of  potential  around  an  electric  double,  that  is, 


ELECTROMAGNETIC  RADIATION 


285 


around  two  spheres  given  equal  but  opposite  charges.     Referring 
to  Fig.  148,  since 

dV  =  -  Rdr, 

V=  -fRdr  =  -J^r  =  f. 
The  potential  at  P  is  (Fig.  149) : 


FIG.  149. 


It  is 


when  r  is  large  compared  with  dZ  (see  note). 
NOTE. — Proof: 


(10) 


a  (q 

^r 


3/ 


9  cos  6 


and 


(2 


by  the  use  of  the  binomial  theorem  it  is  easily  seen  that  this  becomes: 

2qdz  cos  6 
-^2— 

Equation  (10)  may  be  written 


where  /i,  one-half  of  the  length  of  the  oscillator  is  substituted 
for  dz. 

NOTE.  —  Equation  11  is  not  limited  to  spheres  but  is  quite  general  as  long 
as  the  distances  dealt  with  are  long  compared  with  the  length  of  the  oscil- 


286  ELECTRICAL  ENGINEERING 

lator.  Suppose,  for  instance,  that  we  are  dealing  with  a  linear  oscillator. 
We  assume  then  that  the  potential  at  a  point  P  can  be  expressed  as  due  to 
two  point  charges  located  at  some  points  on  each  rod  (not  the  end  of  the 
oscillator)  which  will  give  the  same  potential  as  the  linear  conductor  actually 
gives  at  distances  far  away  from  the  oscillator.  While  this  assumption  is 
quite  justified  when  dealing  with  points  in  space  far  away  from  the  oscillator, 
it  is  obviously  not  at  all  permissible  at  points  near  the  oscillator,  because 
it  is  readily  seen  that  the  potential  distribution  at  the  surface  of  the  two 
halves  of  the  oscillator  must  be  such  that  the  surfaces  themselves  are  equi- 
potential  surfaces  and  two  point  charges,  no  matter  where  located,  can  not 
give  such  equipotential  surfaces.  Fortunately,  we  are  for  practical  purposes 
interested  in  only  what  happens  far  away  from  the  oscillator,  where  equation 
11  applies.  The  subsequent  equations  can  indeed  be  used  with  such  linear 
oscillator  if  instead  of  letting  Q  or  7  represent  the  charge  and  current  respect- 

2  2 

ively,  we  use  the  average  value  along  the  oscillator  which  is  -  Q  and  -  7. 

7T  7T 

The  ratio  between  X  the  wave  length  and  h  the  height  of  the  sending  antenna 
is  in  such  case,  theoretically  4,  but  in  reality  due  to  various  effect  nearer  4.8. 

When  P  is  far  away  from  the  oscillator  the  electric  condition 
is  not  due  to  the  instantaneous  value  of  the  charge  q  at  the 
oscillator  but  due  to  the  value  of  q  which  existed  somewhat 
earlier  in  time. 

Thus  the  charge  causing  the  electric  field  at  P  is  not  q  =  Q  sin  ut 
but  q  =  Q  sin  u(t  —  At)  where  At  is  the  time  required  for  the 
distribution  to  reach  P. 

If  v  is  the  velocity  of  the  propagation  which  is  that  of  light, 
then 

T 

vAt  =  r  'or  At  =  -• 
v 


.'.  q  =  Q  sin  f  ut j 

If  X  is  the  wave  length  then 


27T 


.*.  q  =  Q  sin  I  cot  — —  rj   =  Q  sin  (oot  —  mr)  =  —  Q  sin  (mr  —  cot) 

.'.  V  =  -  2Qh  ^ 

dz  r 

.  _  dF  62     sin  (mr  -  ut) 

(13) 


v  dV  d2     sin  (mr  - 

' 


ay  dydz 


ELECTROMAGNETIC  RADIATION 


287 


H       M 


Z  the  component  of  the  electric  field  intensity  perpendicular  to 
the  x-y  plane  cannot  be  obtained  from  V  alone  as  discussed 
above. 

We  shall  now  consider  some  of 
the  properties  of  the  magnetic  field 
intensities. 

Consider  the  x-y  plane  (Fig.  150). 
It  is  obvious  that  since  the  lines 
of  force  are  circles,  the  sum  of  the 
projections  of  the  components  of 
the  magnetic  field  intensities  along 
the  x  and  y  axes  on  a  radius  vector 
must  be  zero.  Let  L  and  M  be  the  components  of  H  along  the 


FIG.  150. 


x  and  y  axes, 
we  have, 

but 
and 


or 


Since  L  itself  is  negative  in  the  position  shown, 


L  cos  a  +  M  sin  a.  =  0 


cos  a  =  - 
P 

sin  a  =  - 
P 

'.  Lx  +  My  =  0, 

-  -  but  x2  +  y2  = 
x 


L 
M 


.'.  xdx  +  ydy  =  0. 

Thus  L_       dx 

M       dy 

or  Ldy  —  Mdx  =  0 

This  is  satisfied  as  long  as 

L  = 


du  du 

—  and  M  =  —  — 
dy  dx 


(14) 


(15) 


where  u  is  any  function  of  x  and  y 

N  the  component  along  the  2-axis  is  obviously  zero. 

From  equations  (9)  and  (15) 


dt 
~dt 
"aT 


dy 
dL 
dz 


dz 


dz         dxdz 
dL  _    d2u 
dz    ~~~  dydz 
_  dM        dL  _          /dzu        d*u\ 

''  ~dx  "~  fry  " 


dx 


(16) 


288  ELECTRICAL  ENGINEERING 

Referring  now  to  (13)  and  differentiating  X  with  respect  to  t 

—  -  <2Qh  -  (  d*     sin  (mr  ~  w' 
dt    "  2Qhdt(dxdz  '      ~~r 

It  is  evident  by  comparing  (16)  and  (17)  that, 

d          sin  (mr  —  wQ 


Substituting  this  value  in  (16)  we  get: 

dX         .d3      (nsM.  sin  (mr  — 


dt         dtdxdz 

or  v       on, 

A  =  ZQfi 


F  =  2Q/i 


N  =  0 
where  n  =  sin  (mr  -  cop 


It  is  now  a  simple  matter  to  get  the  different  derivatives  of 
n.  An  inspection  of  the  several  terms  will  readily  show  that  some 
are  much  larger  than  others.  There  is  little  object  in  investi- 
gating conditions  close  to  the  oscillator  by  these  equations  even 
if  all  terms  are  used  without  considerable  caution,  because  an 
approximation  was  made  in  the  assumption  that  the  electric 
field  emanated  from  two  point  charges. 

The  derivatives  contain  trigonometric  terms  having  coefficients 
of  raV2  mr  and  unity. 

The  terms  containing  w2r2  are  so  much  larger  than  terms 
involving  mr  and  unity  that  the  latter  can  be  neglected.  Making 
these  approximations  and  placing  P  in  the  x-z  plane  we  get 
for  distances  involving  several  wave  lengths, 


ELECTROMAGNETIC  RADIATION 


289 


7  =  0 

Z  =  -  --  sin  (wr  — 

7,  =  0 

„..        rr 

M  =  //  = 


sin2  0  . 


. 

sin  (rar  —  cot)  sin 
v 


(19) 


(20) 


Everything  is  given  in  electro-static  units  at  present  since  all 
terms  involve  Q  the  charge  which  is  expressed  in  such  units. 


FIG.  151. 

R  the  intensity  along  the  surface  of  a  sphere  through  r  is 
(from  Fig.  151): 

R  =  Z  sin  B  —  X  cos  6  = -  sin  (mr  —  at)  sin  0  (sin2  0  +  cos2  0) 


or 


R  = 


sin  (mr  —  cot)  sin  0 


(21) 


It  is  of  interest  to  compare  equations  (20)  and  (21) 

47T2 


since 

and 

We  can  write 


X2 


2ir 

ma)  =  - 


47T2 

X2T 


H  =  R 


(22) 


when  the  charge  is  given  in  electro-static  units. 

The  electric  and  magnetic  intensities  perpendicular  to  each 
other  in  space  are  in  time  phase.  Thus  the  product  of  the  two 
represents  power.  (This  is  the  case  only  at  some  distance  from 
the  oscillator,  near  the  oscillator  the  large  part  of  the  fields  is 
in  quadrature.) 


290 


ELECTRICAL  ENGINEERING 


It  is  remembered  that  in  the  ordinary  electric  circuit  involv- 
ing capacity  and  inductance  the  magnetic  and  electric  field 
intensities  are  in  time  quadrature  and,  therefore,  the  product 
represents  "  wattless  power  or  better  reactive  power." 

Energy  Radiated. — Equations  (20)  and  (21)  can  be  trans- 
formed to  read, 


and 


2Ihm   .     , 
H  =  -      —  sin  (mr  —  co/)  sin  0 


21  hm2    .     , 

K  — sin  (mr  —  co/)   sin  6 

co      r 


(22) 


Since 


dq 


q  =  Q  sin  (co/  —  mr)  and  i  =  -37  =  Qco   cos  (co/  —  mr) 


dt 


.'.  I  =  QcoorQ  — 


=  Qco  cos  (mr  —  co/) 


If  7  is  expressed  in  amperes  and  R  in  volts  per  centimeter, 
then 


and 
But 


H  =  0.21  —  sin  (mr  -  co/)  sin  6 


21  Hm2 

R  =  300  X  10  -          -  sin  (mr  -  co/)  sin  6 


m  =  —  and  co  =  2irf  =  2?r  -r-     .', 

A  A 

0.47T/  h 

H  =  -      -  -  sin  (mr  —  co/)   sin  6 
r      A 

1207T/   h      . 

R  =  -     — r-  sin  (mr  —  co/)  sin  6 
r       \ 


(23) 


(24) 


From  what  has  been  shown,  it  is  remembered  that  H  and  R 
are  perpendicular  to  each  other  in  space. 

The  e.m.f.  in  a  circuit  is  proportional  to  R,  the  current  is 
proportional  to  H  and  the  power  to  HR  sin  a,  where  a  is  the  angle 
between  H  and  R. 

Since  these  fields  are  perpendicular  to  each  other  in  space  the 
energy  radiated  in  time  dt  eidt  is  proportional  to  HR,  or,  W  = 
kHRdt  and  it  remains  to  determine  the  value  of  k. 


ELECTROMAGNETIC  RADIATION 


291 


The  voltage  per  centimeter  is  R\  thus  e  =  R  when  considering 
1   cm.   of  circuit.     The  m.m.f.  that  produces  a   magnetic  in- 

0  4:iri 
tensity  H  is  - —  where  I  is  the  length  of  the  magnetic  circuit 


in  air. 


.    .        HI 

~    0.47T 

or,  if  we  consider  1  cm.  length  of  magnetic  circuit, 


i  = 


H 

0.47T* 


Thus  the  energy  transmitted  through  a  square  centimeter  area 
is: 

RHdt 


W  =  eidt  = 


0.47T 


Thus  the  energy  radiated  through  the  whole  sphere  of  radius  r 
enclosing  the  oscillator  is  (from  Fig.  152) : 


FIG.  152. 


s*e  =  TT  rt  =  T  r>Tj 
W  =   I  7r-r-  2wr 

Je  =  0     Jr  =  0     0.47T 


sin  Brdddt 


h 


240  7T2      I2  sin2  (mr  -  ut)  sin3  OdOdt 


but 


-  o 


sin2  (mr  —  <*t)dt  =  -~  approximately 


19 


(25) 


292  ELECTRICAL  ENGINEERING 

and 


sm°  vav  =  y% 

:.  w  =  1600^ 

W  h2 

.'.  watts  =  ^  =  1600  ^-2  I2. 

If  I  is  given  in  effective  current  then, 

h2 
watts  =  3200  ^I2  (26) 

In  the  case  of  wireless  transmission  the  radiated  power  corre- 
sponds to  one-half  of  the  area  of  the  sphere  thus, 

h2 
Watts  =  1600  ^-2/2  (27) 

where  I  is  the  effective  value  of  the  current.     The  " Radiation" 
resistance  is  obviously 

R  =  1600  ^  (28) 

It  is  noted  that  in  the  case  of  wireless  telegraphy  the  energy 
radiated  is  greatest  along  the  equatorial  plane,  that  is,  near  the 
surface  of  the  earth. 

Since  the  receiving  antenna  is  near  the  earth  this  result  is, 
of  course,  very  desirable. 

MARCONI'S  improvement  upon  HERTZ'S  oscillator  resulted 
from  his  connecting  the  lower  end  of  his  oscillator  through  a 
spark  gap  to  ground,  by  which  he  was  able  not  only  to  obtain 
the  maximum  energy,  where  it  was  most  useful,  but  also  to  make 
use  of  half  the  length  of  oscillator  for  the  same  distribution  of 
the  magnetic  and  electric  field  above  ground.  This  will  be  evi- 
dent at  once  if  it  is  considered  that  the  earth  being  a  perfect  con- 
ductor, its  surface  is  an  equipotential  surface. 

It  is  easily  proven  from  the  equations  given  that  the  energy 
received  near  the  surface  of  the  earth  through  unit  surface  is 
1.5  times  the  average  value  of  the  energy  per  unit  surface. 

It  is  also  of  interest  to  note  that  with  an  " ideal"    simple 

antenna  where  X  =  4h  and  the  current  is  zero  at  the  top  at  all 

2 
times  and  therefore  the  average  value  of  the  current  is  - 1  that 

7T 

the  power  radiated  in  watts  is  40/e2  or  the  radiation  resistance 
is  Rr  =  40  ohms. 


ELECTROMAGNETIC  RADIATION  293 

In  this  connection  it  is  of  interest  to  add  that  MAXWELL'S 
general  equation  of  propagation  of  electromagnetic  waves  in 
space  free  from  electric  charges  or  magnets  has  been  shown  to  be : 


where  u  is  any  of  the  components  of  the  electric  and  magnetic 
intensities. 

In  the  case  of  spherical  waves  it  is  readily  proven  by  trans- 
forming the  equation  in  spherical  coordinates  that  any  function 
of  r  —  vt  divided  by  r  satisfies  the  equation.  Thus 

r 
The  function  used  so  far  was 

_  sin  (mr  —  coQ 
r 

which  satisfies  the  above  since  mr  —  ut  =  r =  m(r  —  vt). 

m 

In  the  case  of  sustained  oscillations  the  function  chosen  was 
obviously  most  suitable.  In  the  case  of  damped  oscillations  we 
would  naturally  choose 

TT  =  ~e-a(ut-mr}  sm(mr  -  ut) 
r 

where  A  and  a  depend  upon  the  amplitude  and  damping  of  the 
circuit. 

Of  special  interest  is  the  magnetic  intensity  //  near  the  surface 
of  the  ground  and  the  electric  intensity  R  perpendicular  to  the 
surface  but  near  the  ground. 

Equation  (21)  gives, 

R  = sin  (mr  —  wt)  sin  0  =  2h~ '    -  sin  (mr  —  ut)  sin  0  = 

r  co    r 

4?r  -   /  -  sin   (mr  —  cot)  sin  0 
r       A 

where  I  is  expressed  in  electro-static  units. 

If  the  current  be  expressed  in  amperes  and  the  potential 
gradient  in  volts  per  centimeter 

4wlh       V 

X  T7*  X  300  sin  (mr  —  ojt)  sin  0 
10 

=  377  -  £  sin  (mr  -  ut)  sin  0  (29) 
r  A 


294  ELECTRICAL  ENGINEERING 

Thus  the  maximum  value  of  the  potential  gradient  near  the 
surface  of  the  ground  is 

Rmao.  =  377  -  r-  volts  per  cm.  (30) 

/     A 

If,  therefore,  the  height  of  the  receiving  antenna  is  hi  cm.  the 
maximum  value  of  the  potential  difference  between  earth  and 
top  is 

F         377  h  h  I 

El  =  T~  x  hj 

or  the  impedance  of  the  receiving  antenna  is 
Z      Ei       377  h  , 

2  ~-  7-     ~7~  x  *  onms-  (31) 

and  the  effective  value  of  the  voltage  is  : 

E.  =  IfZ1  (32) 

where  Ef  is  the  effective  value  of  the  voltage  across  the  receiving 
antenna  and  Je  is  the  effective  current  in  the  sending  antenna. 
In  a  simple  antenna  the  current  is  a  maximum  at  the  gap  and  is 
zero  at  the  top.  Thus  the  current  is  not  uniform  as  is  the  case  in 

the  HERTZ  oscillator. 

2 
The  average  value  of  the  current  is  -  7.     With  such  simple 

7T 

antenna  the  wave  length  X  should  be  4/i  if  there  were  no  disturb- 
ing effects. 

Substituting  these  values  we  get  as  the  impedance  of  the  receiv- 
ing antenna  in  an  ideal  simple  antenna 

Z\  =  60  ~  ohms  (33) 

The  magnetic  field  intensity  H  (equation  10)  is  similarly  modified 
to 

H  =*  2  -h—  sin  (mr  —  ut)  sin  B  =  -      r-  sin  (mr  —  wt)  sin  6 

0)      T  T      A 

where  I  is  in  abamperes,  or  if  /  is  expressed  in  amperes  rather 
than  abamperes 


h    .     .  .    .  /Q  ., 

H  =  -    —  r-  sin  (mr  —  cot)  sin  8  (34) 

T        A 

It  is  of  interest  to  note  that  this  agrees  with  the  intensity  due 
to  an  infinitely  long  conductor  if 

h  -   JL 

X    ~    27T* 


ELECTROMAGNETIC  RADIATION  295 

If  the  sending  antenna  were  a  simple  rod  then  the  current  at 
the  top  would  always  be  zero  and  the  average  value  of  the  current 

2 

would  be  -  7.     In  that  case  the  wave  length  would  be  4/i. 

7T 

Substituting  these  values  we  get  : 

0.27 

H  =  -    -  sin  (mr  —  ut) 

near  the  surface  of  the  earth. 

Thus  the  approximation  sometimes  made  in  writing 

0  27 

H  =  —  -  sin(rar  —  to/)  sin  6 

is  not  very  far  from  right  —  and  is  correct  in  the  case  of  an  "  ideal 
simple  antenna." 

It  should  again  be  emphasized  that  equations  (29)  and  (34) 
give  the  values  of  the  electric  and  the  magnetic  intensities  several 
wave  lengths  away  from  the  oscillator. 

It  can  very  readily  be  proven  by  carrying  out  the  differentia- 
tions in  equation  (18)  that  near  the  oscillator  the  magnetic 
intensity  decreases  inversely  as  the  square  of  the  distance  and  the 
electric  intensity  inversely  as  the  cube  of  the  distance. 

Power  Factor  and  Logarithmic  Decrement.  —  Prior  to  the  use 
of  high-frequency  alternators  for  the  production  of  radiation  the 
trains  of  waves  were  oscillating,  with  decaying  current  and  e.m.f. 
in  the  antenna  and  the  word  decrement  had  therefore  a  very 
significant  meaning. 

When  alternators  or  oscillating  arcs  are  used  the  current  and  the 
e.m.f.  at  the  antenna  are  sustained,  and  therefore  "  decrement" 
ceases  to  have  any  meaning. 

It  is,  therefore,  appropriate  to  discuss  the  power  factor  rather 
than  to  try  to  treat  of  the  decrement  in  such  circuits. 

If  RQ  is  the  sum  of  the  radiation  resistance  and  the  effective 
resistance  of  the  wires  and  the  ground  connection,  then  the  power 
consumed  in  the  circuit  is  P  -=  PR0,  where  7  is  the  effective  cur- 
rent. If  E  is  the  effective  voltage,  then 


I  =  2irfCE  =  coCE  where  C  is  given  in  farads, 
thuS  Pf  =  a>CR0  (35) 


296  ELECTRICAL  ENGINEERING 

Numerical  application : 
Let  C  =  003  m-f.  -  ~i  farads. 

h  =  50  m. 
X  =  3000  m. 

.'.    Radiation    resistance  =  1600  (^7.7^;) 2  =  0.5   ohm  approxi- 

,  XoUUU/ 

mately 

W    =    27T/   =    27T—     =    27T105. 

A 

Let  the  effective  resistance  of  the  wires  and  ground  be  2  ohms, 
then 

R0  =  2.5  ohms. 

.'.  Pf  =  27rl05  j^  2.5  -  0.0047 

or  approximately  one-half  of  1  per  cent. 

The  radiated  energy  corresponds  in  this  case  of  course  to  only 
one-fifth  of  this  amount. 

The  product  of  the  current  and  the  e.m.f.  is  200  times  as  great 
as  the  power  consumed  in  heat  and  radiations  and  1000  times 
as  great  as  the  power  radiated. 

Determination  of  the  "Logarithmic  Decrement." — If  a  con- 
denser is  discharged  in  a  circuit  of  negligible  resistance  an  alter- 
nating current  will  flow  indefinitely,  and  no  energy  will  be  ex- 
pended since  the  energy  is  transferred  alternately  between  the 
magnetic  and  the  electric  field. 

When  the  current  is  a  maximum  (either  positive  or  negative) 
the  e.m.f.  across  the  condenser  is  zero;  when  on  the  other  hand 
the  current  is  zero,  the  e.m.f.  is  a  maximum. 

Thus  twice  in  each  cycle  the  magnetic  energy 

wm  = 

is  transferred  to  electric  energy 

We  = 

The  total  amount  of  energy  in  joules  surging  during  a  cycle 
is  then 

W  =  L/2  where  /  is  the  maximum  value  of  the  current, 
or, 

W  =  CE2  where  E  is  the  maximum  value  of  the  voltage. 


ELECTROMAGNETIC  RADIATION  297 

If,  however,  the  circuit  contains  resistance,  'the  current  will 
not  alternate  indefinitely  but  will  die  down  gradually,  the  rate 
of  decay  being  greater  the  greater  the  energy  consumption. 
During  these  oscillations  energy  is  also  transferred  between  the 
electric  and  magnetic  field  but  each  pulse  of  energy  is  smaller, 
than  the  preceding  by  the  loss  of  energy  in  the  resistance. 

Ultimately  all  energy  stored  in  the  condenser  becomes  dissi- 
pated in  heat  or  radiated  away. 

The  energy  stored  in  a  condenser  is  %CE2  joules  where  E 
is  the  voltage  and  C  the  capacity  in  farads.  Thus  if  the  condenser 
is  charged  and  discharged  N  times  per  sec.  the  sum  of  the  energy 

N 
converted  to  heat  and  radiated  away  is  -^CE2  joules  per   sec. 

Zi 

or  watts. 


FIG.  153. 

Thus  Wi  =  ^CE2  watts  (36) 

z 

In  a  circuit  of  concentrated  inductance  and  capacity  it  is 
shown  in  the  elementary  theory  of  alternating  current  that  the 
oscillating  current  can  be  expressed  quite  accurately  by  the 
following  equation: 

i  =  I6-atsmut. 

IT* 

Where  a  =  ^  and  RQ  and  L  are  assumed  constants  which  how- 

£iLi 

ever  is  not  the  case  in  ironclad  inductors  and  in  arcs. 


298  ELECTRICAL  ENGINEERING 

The  ratio 


g 


as  is  readily  proven. 

The  logarithmic  decrement  is 


(37) 


Incidentally  5  is  also  the  ratio  between  the  energy  absorbed  by 
the  resistance  and  the  surging  energy  per  cycle. 


Thus  I2RoT  R0T 

-~ 


which  agrees  with  (37). 

In  the  case  of  the  HERTZ  oscillator  or  the  umbrella  type  of 
antenna  the  inductance  is  confined  largely  to  the  linear  con- 
ductor and  the  capacity  to  the  spheres  or  superstructure;  thus 
we  may  consider  the  inductance  and  capacity  as  separated  rather 
than  distributed,  thus 

(38) 


(39) 

The  resistance  in  the  above  formula  is  the  sum  of  the  radiation 
resistance,  the  resistance  of  the  wires  (taking  into  consideration 
the  skin  effect),  the  ground  and  the  radiation  resistance. 

When  an  arc  is  used  the  resistance  of  the  arc  should  also  enter. 
Unfortunately  the  latter  is  not  a  constant  but  depends  upon  the 
current  carried,  and  hence  the  decrement  is  not  logarithmic. 
However,  for  the  purpose  of  this  article  the  arc  resistance  may 
be  assumed  constant  at  say  5  ohms.  For  a  very  complete  dis- 
cussion of  this  whole  subject  the  reader  is  referred  to  FLEMMING'S 
"  Principles  of  Electric  Wave  Telegraphy." 

Equation  39  contains  the  inductance  and  capacity  as  well  as 
the  resistance.  The  inductance  is  usually  very  difficult  to 
determine  since  at  different  wave  lengths  more  or  less  inductance 
is  added  to  that  of  the  antenna  proper.  The  capacity  of  the 
antenna  is  however,  usually  not  changed  but  it  depends  upon  the 
construction  of  the  aereal.  The  complexity  of  the  structure  is, 
however,  such  that  its  value  can  hardly  be  calculated  except  in 
the  very  simplest  cases  —  rarely  used  in  practice. 


ELECTROMAGNETIC  RADIATION  299 

FLEMMING  expresses  the  approximate  capacity  of  a  vertical 
wire  of  radius,  h  cm.  long  as: 

Cv  =  -      — - —  —  farads, 

2  log  -  X  9  X  1011 

when,  as  is  the  case  in  wireless  stations  the  lower  end  of  the  wire 
is  near  ground,  the  capacity  may,  however,  be  say  10  per  cent, 
greater. 

He  also  expresses  the  capacity  of  a  horizontal  wire  placed  h\ 
above  ground  as 

I 


Ch  = 


2  log—-1  X  .9  X  1011 


where  I  is  the  length  in  centimeters  and  h  the  height  above 
ground. 

Thus  the  capacity  of  a  T-shaped  antenna  may  be  approxi- 
mated as: 

c  =  cv  +  ch 

obviously  the  total  capacity  is  not  at  all  proportional  to  the 
number  of  wires  connected  in  multiple.  It  is  only  slightly 
increased  as  the  number  of  wires  is  increased. 

If  the  value  of  the  capacity  is  difficult  to  calculate  accurately 
it  is  measured  relatively  easily  and  will  therefore  be  assumed 
as  known.  It  ranges  according  to  ZENNECK  approximately  as 
follows : 

0.001  m-f.  in  torpedo  boat  antenna. 
0.002  m-f.  in  battleship  antenna. 
0.007  m-f.  in  BRANTROCK  station. 
0.18    m-f.  in  NAUEN  high-power  station. 

The  capacity  of  the  antenna  of  the  experimental  installation  at 
Union  College  is  0.0012  m-f. 

When  the  wave  length  is  considerably  more  than  four  times 
the  height  of  the  antenna  the  current  distribution  is  fairly  uni- 
form in  the  conductor,  and,  the  circuit  can  be  treated  as  consist- 
ing of  "bunched"  rather  than  distributed  inductance  and 
capacity  when  the  following  relation  obtains. 

T  =  27T/LC     /.  L  =      ~ 


300  ELECTRICAL  ENGINEERING 

thus  .C 


-f  (40) 

\ 

Numerical  application:  Union  College  station  with  an  antenna 
having  a  capacity  of  0.0012  m-f.  sending  out  waves  of  700  m. 
length.  Assume  R0  =  10  ohms.  (By  far  the  greater  part  of 
this  is  the  ground  and  spark  resistance.) 


Then  5  =  2o10_!2_  3  X  _ 

°1010     70,000 

In  the  case  of  the  simple  antenna  it  has  been  shown  that  the 

2 
radiation  resistance  assuming  X  =  4/i  and  Iavo>  =  -  /  is  40  ohms. 


Thus  the  radiation  decrement  is  : 


3  X  1010  = 


2  log     9  X  1010       log 

In  reality  due  to  the  proximity  of  the  earth  and  other  causes 

the  wave  length  is  nearer  4.8  than  four  times  the  antenna  height, 

2 
and  the  average  value  of  the  current  is  nearer  0.7  and  -. 

7T 

Substituting  these  values  we  get : 

R  =  34  ohms  instead  of  40  ohms 
and  the  radiation  decrement  for  the  simple  antenna  is : 

,_*!!*  8  X16»*-^  (41) 

2  log  ^9X10"'       log" 

Abraham  gives  s        2.45 

o  —  -      r* 
i       » 
]°g  ~ 

General    Conclusions. — Since    the    power   radiated    from    an 
antenna  is: 

W  =  1600  ^  72 

it  is  evident  that  at  a  given  voltage  as  the  capacity  of  the  super- 
structure is  increased  the  current  and  the  wave  length  are  in- 
creased. Since,  however,  the  energy  is  proportional  to  the  square 


ELECTROMAGNETIC  RADIATION  301 

of  the  current  and  the  wave  length  is  proportional  to  -\/C  it 
follows  that  by  adding  capacity  to  the  superstructure  and  there- 
fore increasing  the  wave  length  the  radiated  energy  is  increased. 

Therefore,  if  the  capacity  is  made  four  times  as  great,  the  cur- 
rent2 is  16  times  as  great  and  X2  is  only  four  times  as  great,  and 
hence,  the  radiated  energy  for  the  same  antenna  height  is  in- 
creased four-fold. 

Unfortunately,  however,  there  is  hardly  a  practical  way  of 
increasing  the  top  capacity  without  decreasing  the  effective 
height  so  that  the  gain  is  not  as  great  as  indicated  and  if  the 
umbrella  is  carried  to  an  extreme,  the  effective  height  may  be  so 
much  decreased  that  the  energy  radiated  may  eventually  begin 
to  decrease. 

With  a  given  construction  of  the  antenna  the  wave  length  may 
be  increased  by  the  introduction  of  inductance.  In  this  case  the 
energy  radiated  is,  however,  reduced. 

It  is  noted  that  for  a  given  current  the  radiated  power  is 
greater  the  higher  the  frequency.  This  does,  however,  not 
necessarily  mean  that  the  power  received  is  greater,  since  as  will 
be  shown  later  the  absorption  of  energy  in  space  is  much  greater 
with  short  wave  length  than  with  long. 

At  times  it  is  necessary  to  send  at  two  widely  different  fre- 
quencies. The  natural  wave  length  may  be  say  600  m.  and  it  is 
desired  to  communicate  at  a  wave  length  of  300  m.  In  that  case 
a  condenser  may  be  connected  in  the  series  with  the  antenna. 
Since  two  condensers  in  series  have  a  smaller  capacity  than  each 
and  thus  the  frequency  is  increased. 

The  relation  between  the  effective  value  of  the  antenna  current 
and  the  maximum  instantaneous  value  of  the  current  and  e.m.f. 

If  the  damping  is  not  excessive  the  discharge  current  of  a 
condenser  of  voltage  E  can  be  represented  by  the  following 
equation : 

_^o 

i  =  EuCe    2L  sin  u 

=  Ie~at  sin  co£ 
where 

/  =  EuC  and  «  =  g  =  |- 

RQ  being  the  total  resistance  in  the  circuit  which  is  assumed 
constant,  not  depending  upon  the  current. 


302  ELECTRICAL  ENGINEERING 

The   rate  at  which   energy  is  being  converted  to  heat  and 
radiated  is  then: 

Ro  P€-*at  sin2  coZ. 

The  energy  developed  in  one  train  of  waves  then  is, 


R0I2  e~2at  sin2  utdt  =          °  approximately  (43) 

If  the  antenna  is  charged  and  discharged  N  times  per  sec.  then 
the  power  is 

W    _    A/- ±L°  (AA\ 

43/ 

If  Ic  is  the  effective  value  of  the  antenna  current  as  read  by  a 
hot-wire  instrument, 

.:I  =  Ie^jy  (45) 

and  since  / 


Substituting  JRo 


and  r          1 

L  = 


we  get  l2R0C 


Je      ,„„„  (47) 


These  equations  connect  the  instantaneous  max.  values  of  the 
antenna  current  and  e.m.f.  with  the  effective  current  read  by  a 
hot-wire  instrument. 

Numerical  application :    At  the  Union  College  station  R0  =  10, 

12 

X  =  700  m.,    C  =  -,  N  =  500.     Using  the  small  sending  set. 

I€  =  2.5  amp. 
3  X  1010 


70,000 


0.43  X  106 


ELECTROMAGNETIC  RADIATION 


303 


I  =  2.527T.43106 


=  46  amp. 


E  =  14,400  volts. 

Relation  between  E.m.fs.  Frequencies  and  Coupling  in  In- 
ductively Connected  Circuits. — Let  e\  in  Fig.  154  be  the  voltage 
across  the  primary  capacity,  e2  in  Fig.  154  be  the  voltage  across 
the  secondary  capacity. 

Then  neglecting  resistance  we  get: 


^7    =   0 


but, 


dt 


"  dt 


~dt 


l~df 


-  =  0 


and 


(1) 


(2) 


FIG.  154. 

Substituting  the  current  values  of  equation  (2)  and  equation  (1), 
and  writing 

ei  =  Ei  sin  ut 

ez  =  EI  sin  (ut  +  a). 
Thus 


we  get, 


(3) 
(4) 


304  ELECTRICAL  ENGINEERING 

From  (3) 

61  =  F^rfc^2  (5) 

Substitute  (5)  in  (4)  and  assume  that  C\Li  =  C2L2  =  CL,  that 
is,  assuming  that  the  circuits  when  independent  are  tuned  to  the 
same  wave  lengths,  then, 


(6) 


-  2LCco2  -  co4(L2C2  -  lf2CiC2)  -  0 
1  ±  k 


"  CL(1  - 
where 

7  M 

k  =  . 


(7) 

&nu\j       \  j.   —   t\j~  \  a.    —   A/~ 

or, 


(8) 


where  /o  is  the  frequency  of  each  circuit  when  alone. 

It  is  seen  from  these  equations  that  two  frequencies  exist  in 
the  circuit  and  that  they  become  nearer  and  nearer  alike  as  the 
coefficient  of  coupling  is  decreased,  that  is  the  less  the  value  of 
the  mutual  induction  as  compared  with  the  self-induction. 

In  the  case  of  transformation  by  ordinary  transformer  where 
the  mutual  induction  is  almost  perfect,  only  one  frequency  will 
appear,  namely,  /2  the  forced  frequency  which  in  that  case  is 

/2  =  /0 — -7=-    1°  other  words  the  radiated  frequency  has  only  one 

V2 

value  and  that  value  is  70  per  cent,  of  that  of  each  circuit  when 
alone. 
Since 

/I    _    \2 
/2  Xi' 

It  follows  that  two  different  wave  lengths  are  transmitted 
and  that 

fT^fc 


X2        -    '    '- 


ELECTROMAGNETIC  RADIATION  305 

Wave  meters  are  used  to  show  the  wave  length,  and  hence,  if 
the  wave  length  is  known  the  coefficient  of  coupling  can  be  deter- 
mined, it  is: 

X22  -  Xa1 

(10) 


k  = 


(11) 


A2   -f  AI* 

It  is  evident  from  the  above  that  the  current  and  voltage 
in  two  such  circuits  must  be  expressed  as  functions  of  two 
frequencies. 

Let  61  =  AI  cos  coi£  4~  Bia)2t 

e2  =  A2  cos  coit  -f-  B2  cos  ai2t 

These  equations  are  justified  since  the  resistance  is  negligible, 
and  hence,  no  appreciable  phase  displacements  exist  between 
the  two  voltages. 

For  t  =  0, 

ei  =  Elt 
and 

e2  =  0. 
/.  E!  =  A,  +  B1  | 

0  =  A2  +  B2  \ 

Consider  then  the  two  waves  separately. 
We  have  from  (5) 


(12) 


1  - 


and 


where 


1 


A2 

Bl 


(13) 


a.  = 


ft  = 


Thus  from  (12) 


1  —  jLido?22 

Ai  =  E1-  B1 
A2  =  -  B2 


and 


A,  =  El  -  Bi     .'.  A!  =   - 


aEi 


(14) 


—  a. 


306 

But 
and 


ELECTRICAL  ENGINEERING 
El 


jCl  1  A 

.'.   A<t   = 

a 


—  a 


B»=  -  A*     /.  B2  =  - 


—  a 


(15) 


.'.  Substituting  these  values  in  equation  (11) 

•pi 

[COS  Uit   —   COS  0>2 


—  a 


Thus 


From  (13) 


from  (6) 


-  a 


wi- 


C02 


2MC 


-  a 


(16) 


Equation  (16)  shows  the  relation  between  the  maximum  voltage 
and  the  capacity  when  the  circuits  have  negligible  resistance. 

When  damped  oscillations  are  considered  the  equations  become 
more  involved. 


FIG.  155. 

Consider  the  simplest  case  when  the  primary  of  the  exciting 
transformer,  Fig.  155,  is  supplied  with  power  from  an  alternator 
or  other  source  of  sustained  oscillations. 

Due  to  the  mutual  induction  between  the  primary  and  second- 


ELECTROMAGNETIC  RADIATION  307 

ary  circuits  an  e.m.f.  EI  sin  co^   will   be   impressed   upon   the 
secondary. 

The  differential  equation  of  the  secondary  circuit  is  thus  : 

EI  sin  &it  =  iRz  +  L^rrr  +  ^2 
at 

where  e2  is  the  voltage  across  the  secondary  condenser. 
But  .       ~  de2 


.-.  E1  sin  Wl«  =  C,R        +  C2L22  +  e2  (1) 

The  sine  term  can  be  eliminated  by  two  successive  differentia- 
tions and  the  result  will  be  a  well-known  linear  differential 
equation  of  the  fourth  order  the  solution  of  which  is: 

e2  =  Efsm  (wi$  +  <p)  +  E'e  ~at  sin  (w2£  +  t)  (2) 

The  first  term  shows  the  value  of  the  permanent  voltage  of  the 
secondary  circuit  —  of  primary  frequency,  the  second  that  of  the 
transient  which  very  soon  ceases  to  exist. 

Thus,  if  it  is  desired  to  study  the  constants  of  the  antenna  the 
transient  term  may  be  neglected  and  the  permanent  voltage 
becomes 

e^  =  E  sin  (wi£  -f-  v?). 

Substituting  this  value  in  the  differential  equation  we  get 
after  some  simple  transformations  the  following  relation  between 
the  maximum  value  of  the  secondary  voltage  and  the  induced 
voltage. 


The  secondary  frequency 

/    = 

when  the  circuit  contains  no  resistance  and 


when  the  resistance  is  R2 

Thus  1 


\J2LJ1 

or  1 

77-7-  =     ^2     + 
U  2i-/2 

20 


308  ELECTRICAL  ENGINEERING 

where  R2 

az  =  2LS 


•     T 

" 


co!2  -  o>22  -  a)2  +  (2eoia2)2 

When  the  secondary  circuit  has  the  same  natural  period  as  the 
primary  impressed  frequency  the  secondary  current  becomes  a 
maximum. 

Thus  for 

0>2    =    COi       I    =    Ir 


since  a24  is  very  small  compared  with  4coi2a22. 
This  is  readily  shown  to  be 


.*.  The  square  of  the  effective  value  of  the  secondary  currents  is 


L2    -    C022    -    «22)2 

or22  is  small  compared  with  co22,  thus 
I    _  2a2a>i 

Tr    = 


tW 


//I 

\/(l 

\    \ 


but 


and 

B 

02    — 


and 


"  2L2/2 

•'•  2ZT2  =  52j 


ELECTROMAGNETIC  RADIATION  309 

r*       (-i-Y 

I2  \  irfi  / 


.'.(I  -o;2)2 
where 

* 

or 

_ 
- 

and 


If  x  is  near  unity  then  1  —  x2  =  (1  +  x)  (1  —  x)  =  2(1  —  x) 
and 


=  27r(l   -  x) 


rjL 

\ir2  -  r 


If  the  secondary  current  J  is  read  by  a  hot  wire  instrument 
then  since  the  effective  values  are  proportional  to  the  maximum 
values, 


or, 


er 

If  the  frequency  of  the  secondary  is  so  adjusted  that  J2  =  -£- 

z 

then  we  get 


a  formula  which  is  used  estensively  in  connection  with  wave 
meter  measurements. 

Inductively  Coupled  Oscillating  Circuits  Having  Considerable 
Damping.  —  We  have  shown  (equation  5)  the  following  relation 
between  the  effective  secondary  current  Jr  at  resonance  and  the 
induced  e.m.f.  EQ  when  the  primary  is  supplied  from  a  source  of 
sustained  power 


310  ELECTRICAL  ENGINEERING 

The  corresponding  equation  when  both  the  primary  and  second- 
ary circuits  are  oscillating  has  been  worked  out  by  BJERKNES  and 
others  who  found  that  as  long  as  the  decrements  are  small  the 
following  relation  obtains: 


16/3L 


In  this  equation  Jmax*  is  the  maximum  possible  value  of  cur- 
rent read  by  a  hot  wire  instrument  in  the  antenna  circuit. 

N  is  the  number  of  condenser  discharges  per  second. 

EQ  is  the  maximum  value  of  the  e.m.f.  induced  in  the  antenna 
circuit.  L2  is  the  inductance  of  the  antenna  circuit  in  henrys, 
/  is  the  frequency  and  di  and  dz  the  logarithmic  decrements  in 
the  primary  and  antenna  circuits  per  full  period. 


7i  = 
E0  =  <**MCiEi  and  E£ 


2  _          _  = 

"  16/3  " 


Similarly, 


In  the  case  of  sustained  primary  power. 

The  maximum  instantaneous  value  of  the  antenna  current  is 
from  (45)  remembering  that  in  these  equations  the  decrements 
per  full  period  is  used. 

T          2    A   f  S  T          2 

T    2    "  max.      ^  J  "2    A    "max.     f? 

~w~  i\r      ~A^52 

The  maximum  instantaneous  value  of  the  antenna  voltage  is 

/2 


(ID 


Numerical  Examples.  —  Union  College  small  set. 
El  =  5000  volts 
120  . 

Ci  ==  JQ-TO  farads 


farads 


N  =  500 


ELECTROMAGNETIC  RADIATION 


311 


and 


or 


X  =  700m.;     /.  /  =  0.43  106 
>!  =  0.05  52  =  0.10  k  =  0.10     .'.  did2(d!  +  d2)  =  ~-6 

i  o  v  1 20  1 05 

.2  =  50,000  0.43  106  ^f^      25  106  0.01  ~~  =  20 
10  75 

T          _  jr 

•  •     *J  max.     ~      Tt.tJ 
4    \/   OH 

-  0.43  106  X  0.10  =  1375 


% 


72  =  37  amp. 
1010 


278  X  37  =  11,400  volts. 


2x0.43  106  12 

BJERKNES  has  shown  how  with  a  slight  modification  equation 
(6)  can  be  used  to  determine  the  decrement  of  the  secondary 
circuit  which  may,  for  instance,  be  the  antenna  circuit  by  means 
of  a  third  tuned  circuit  which  is  called  a  wave  meter: 

This  expression  is: 

d  +  5i  =  27r  (l  -  —}  (12) 

where  5  is  the  decrement  of  the  circuit  being  tested  and  <5i  is 
the  decrement  of  the  meter. 

The  formula  is  limited  as  is  the  case  of  equation  (6)  to  the 
condition  that 


Jr  and  J  being  the  effective  values  of  the  current  in  the  wave 
meter. 

It  is  also  limited  to  the  condition 
that  both  d  and  Si  are  small  and 
that  di  is  considerably  smaller  than 
8  and  that  finally  Xi  and  X2  do  not 
differ  more  than,  say,  5  per  cent. 

Referring  to  Fig.  156,  W  is  the 
wave  meter  which  is  a  calebrated 
closed  circuit  of  known  inductance, 
capacity  and  therefore  of  known 
natural  period.  The  resistance  is  FlG-  156- 

made  as  low  as  possible  so  that  the  decrement  of  the  meter  is  small. 

The  value  of  the  current  or  the  (current)2  is  frequently  deter- 
mined by  means  of  a  low  resistance  heating  element,  actually  a 
thermal  couple,  which  supplies  a  direct  current  to  a  galva- 
nometer G. 


312  ELECTRICAL  ENGINEERING 

In  that  case  the  galvanometer  deflection  is  obviously  pro- 
portional to  the  square  of  the  current  value. 

The  procedure  is  as  follows.  The  meter  is  loosely  coupled  to 
the  antenna  and  the  capacity  of  the  wave  meter  is  varied  until 
the  largest  galvanometer  deflection  G>  is  obtained  and  the 
corresponding  wave  length  \o  is  read. 

Then  the  capacity  is  changed  so  that  the  deflection  of  the 

C1 

galvonometer  is  -~  when  the  meter  reads  shorter  wave  length. 

We  have  then  from  (12) 

5    +    dl    =    27T    (l    -    ^]  • 


To  determine  the  decrement  of  the  meter  it  is  desirable  to 
insert  in  the  meter  circuit  such  non-inductive  resistance  that  at 
resonance,  that  is  when  the  wave  meter  reads  Xo,  the  galvanom- 

C* 

eter  deflection  is  ~ 


The  capacity  is  then  varied  until  the  galvanometer  deflection 

C1 

-^  when  the  wave  length  is  X2. 

We  have  then  if  52  is  the  decrement  due  to  the  added  resistance, 


It  has  been  shown  in  equation  (8)  that  the  relation  between 
the  effective  values  of  the  resonance  current  with  different 
decrements  are  related  as  follows: 


J/2          dd*  (dl  +  da) 
In  our  case 


Jr'2        Gr> 

2 

d\  =  d  =  decrement  of  the  antenna. 

d'z  =  di  +  52  =  decrement  of  the  wave  meter  in  second 

test. 

di  =  8  =  decrement  of  the  antenna. 
<22  =  5i  =  decrement  of  the  meter  in  the  first  test. 


ELECTROMAGNETIC  RADIATION  313 

2  =  (g* +  *«)(*  + fr +  frJ 
5i(5  +  52) 

1-^ 

Si  +  52  Xo       Si  +  52 


Si         }        Xi 
Xo 

-   8  =2I  (Xl  ~  Xa)  (x»  ~  x^) 

*  *  \  \         i     \  o\ 

AQ  AO  "T~  *»2   —  ^Ai 

Numerical  Example. — 
X0  =  500  m. 
Xi  =  485  m. 
\z  =  475  m. 

.-.  «  +  «.!  =  2ir  (l  -^)   =  0.189. 


-  ^  (J  -  500)   -  °'314 


-  0.125. 

27T     10  X  25 


=  0. 


500         5 
d  =  0.126. 

Conditions  Affecting  the  Receiving  Station. — It  has  been  shown 
that  at  some  distance  from  the  sending  antenna  the  maximum 
value  of  the  potential  gradient  in  volts  per  centimeter  near  the 
equatorial  plane  is 

G==l20irhI  (1) 

where  /  is  the  maximum  value  of  the  current  at  the  sending 
antenna,  the  current  being  assumed  the  same  at  all  points  of 
the  conductor.     The  dimensions  are  given  in  centimeters. 
A  more  general  formula  would  be 


*  (  . 

E2  =       —j-  (2) 

where  E2  is  the  maximum  value  of  the  voltage  across  the  whole 
receiving  antenna,     a  is  a  correction  factor  for  the  current  dis- 

2 
tribution  which  is  -  for  a  simple  antenna  and  unity  for  an  antenna 

7T 

in  which  the  height  constitutes  only  a  fraction  of  a  quarter  wave, 
as  is  most  frequently  the  case  in  actual  practice. 

h\,  hz,  X  and  r  may  be  given  in  any  units  as  long  as  they  are 
the  same,     hi  and  h2  are  the  heights  of  the  sending  and  receiv- 


314  ELECTRICAL  ENGINEERING 

ing  antenna,  X  the  wave   length  and  r  the  distance  between  hi 
and  hz. 

In  order  to  be  applicable  to  wireless  transmission  this  formula 
needs  to  be  elaborated  in  several  respects. 

(a)  The  voltage  is  actually  greater  due  to  the  concentration 
of  energy  as  the  waves  sweep  over  the  surface  of  the  earth. 

(b)  The  voltage  is  smaller  on  account  of  the  energy  which 
strays  away  from  the  curvature  even  if  the  surface  of  the  earth 
is  assumed  to  be  perfect  of  conductivity. 

(c)  The  voltage  is  reduced  on  account  of  the  energy  absorption 
of  the  earth  current  which  effect  is  prominent  near  the  sending 
conductor  where  the  concentration  of  current  is  greatest. 

(d)  The  voltage  is  sometimes  increased,  but  more  often  re- 
duced, due   to  reflection,   absorption,  etc.,  depending  upon  the 
condition  of  the  atmosphere. 


FIG.  157. 

Conditions  (c)  and  (d)  have  not  been  studied  theoretically, 
but  a  considerable  amount  of  data  has  been  given  from  actual 
tests,  notably  by  AUSTIN  and  FULLER.  l 

The  Effect  of  the  Curvature  of  the  Earth. — Assume  that  the 
sending  antenna  is  at  A  and  the  receiving  antenna  at  B,  Fig. 
157. 

The  distance  between  A  and  B  is  -=-.     In  the  case  of  a  plane 

z 

wave  the  receiving  antenna  for  the  same  distance  would  then  be 
at  C  where, 

A  -C  =  \R. 

Thus  in  this  latter  case  the  energy  would  be  spread  over  a 

AUSTIN,  Bulletin,  Bureau  Standards,  1914. 
FULLER,  Proc.,  A.  I.  E.  E.,  April,  1915. 


ELECTROMAGNETIC  RADIATION 


315 


circumference 


wnereas  due  to  the  curvature  of  the  earth 


the  circumference  is  only  2irR.     There  is,  therefore,  a  concentra- 
tion of  energy  which  can  be  represented  by  a  coefficient 


k'  = 


and  since  the  intensity  of  the  electric  field  is  proportional  to  the 
Venergy,  the  concentration  coefficient  for  the  electric  field  at  a 
distance  r  under  the  condition  given  above  is 


R0.     .'.  Energy 


Let  distance  AC,  Fig.  157,  be  equal  to  AB 
per  unit  length  of  circumference  at  C  is 

E 


Energy  per  unit  circumference  at  B  is 

_E_ 
2irR  sin  6 

2irR6  6 


2irR  sin  0       sin  B 


or      = 


(3) 


FIG.  158. 

The  effect  of  the  straying  of  power  on  the  potential  gradient 
due  to  the  curvature  of  the  earth  is  included  in  the  equation 
according  to  theoretical  works  done  by  SUMMERFIELD  and  ZEN- 
NECK  by  the  introduction  of  a  divergence  factor. 

0.0019r 


316  ELECTRICAL  ENGINEERING 

AUSTIN'S  experiments  indicate,  however,  that  with  continuous 
waves  this  coefficient  is: 

0.0915r 


and  FULLER'S  experiments  show 


0.0045r 

=    € 


(4) 


AUSTIN'S  equation  gives  values  which  lie  between  ZENNECK'S 
and  FULLER'S  and  has  the  advantage  of  being  simpler  than  the 
other  two. 

Thus  the  general  formula  for  continuous  waves  becomes  : 


,  , 

#2  =  kki  --  —  (5) 

Note,  however,  that  in  equation  (4)  the  dimensions  are  ex- 
pressed in  kilometers. 

The  maximum  value  of  the  antenna  current  in  the  case  of 

Tjl 

sustained  oscillations  is  evidently  ^2  —  TT  where  R%  is  the  total 

/l2 

resistance  of  the  antenna  that  is  the  radiation  resistance,  the 
effective  resistance,  ground  resistance,  and  resistance  of  the 
receiving  device. 


The  equation  01  the  current  in  the  case  of  damped  oscillations 
is  slightly  different. 

It  has  been  shown  that  if  an  e.m.f.,  EQ,  is  impressed  on  a  tuned 
circuit  the  following  relations  obtain: 

T   2    = 

~ 

where  EQ  is  the  voltage  induced,  which  in  our  case  is  E%.  di  and 
d-2,  are  the  decrements  in  the  two  circuits. 

Thus  di  and  d%  are  in  this  case  the  decrements  of  the  sending 
and  receiving  circuits  respectively. 

Equation  (7)  may  be  written: 


/        di\ 

,«,*•  ^i  +  ^ 

But  the  decrement  of  the  receiving  antenna  is 


2-L2/ 


ELECTROMAGNETIC  RADIATION  317 

where  /     M20     h 

= 


.      ,  = 


4/di 

wnere  j  i  is  in 

Tt2   Nm 

g*Ji' 


but  /i2  =  Ji2  —  ^  where  Ji  is  the  effective  value  of  the  sending 
antenna  current. 


4/fl22(l+JW        RS    (l+J) 

\  «2/  \  «2/ 

and  J2  =  _   ^/^  (g) 


The  effective  value  of  the  voltage  across  the  receiving  antenna 

TT  ^  AsaJi  (10) 


_ 

where  €2  is  the  effective  value  of  the  receiving  voltage  and  Ji 
is  the  effective  current  at  the  base  of  the  sending  antenna. 

It  is  evident  from  the  above  that  the  ratio  between  the  effect- 
ive values  of  the  received  e.m.f.  with  sustained  and  with  damped 
oscillations  is: 


damped 

if  the  decrements  of  the  sending  and  receiving  antennas  were  the 
same  then  the  ratio  would  be  V2. 

Method  of  Determining  Power  Received.  —  AUSTIN  based  his 
determinations  on  the  fact  that  if  in  two  circuits  in  parallel  we 
know  the  power  in  one  we  can  calculate  the  power  in  the  other 
and  the  total  power  from  the  relations  of  the  resistances  Rys' 
in  the  circuits. 

The  total  power  supplied  is 

FT       F(E  +  E\       E»R  +  S- 
~  E  \R  +  S)~          RS 

The  power  of  circuit  R  is 

P    -^ 
Fr  ~  R 

•   %l  -  R  +  S     P  -  R  +  S 
"  Pr~      RS  S 


D      I       Cf 

or  the  total  power  =  —  «  —  Pr. 


318  ELECTRICAL  ENGINEERING 

The  minimum  power  Pr  required  for  distinguishing  between  dots 
and  dashes  of  resistance  R  is  determined  experimentally  by  ob- 
serving the  current  in  the  receiving  antenna  under  conditions 
that  can  be  conveniently  controlled. 

Knowing  Pr  and  R  and  the  resistance  S  which  is  shunted  across 
the  telephone  receiver  enables  one  to  determine  the  total  power 
received.  In  FULLER'S  experiments  at  Honolulu  this  minimum 
power  was  found  to  be  3.2  X  1010  watt,  when  dealing  with  sus- 
tained oscillations. 


APPENDIX  I 

Partial  Differentiation.  —  The  complete  differential  of  a  func- 
tion V  of  several  independent  variables  r,  <p,  6  is  recalled  to  be: 

"-£*+£*+S« 

In  words  this  equation  reads:  The  total  differential  of  V  is 
the  sum  of  the  partial  differentials  of  V  with  respect  to  the 

independent  variables.      —  meaning  the  derivative  of  V  with 

respect  to  r  when  <p  and  6  are  considered  constant. 

If  the  independent  variables  r,  <p,  and  6  are  some  functions  of 
a  single  other  variables  t  the  derivative  of  V  with  respect  to  t 
is  obtained  by  simply  dividing  equation  (1)  by  dt. 

Thus:  dV      dV  dr      aV  d<p      dV  dd 

dt     ~  dr    dt  +  d<p    dt  +  d6    dt 

If  the  independent  variables  r,  <p  and  6  are  functions  of  several 
other  independent  variables,  for  instance  x,  y,  z,  then  the  partial 
derivative  of  V  with  respect  to  x  is  obtained  in  a  similar  way  by 
dividing  the  equation  by  dx,  remembering,  however,  that  now 

-j—  is  the  partial  derivative  and  should  be  written  -r  —  •• 
dx  dx 

Thus         ^  =  iZ^,^I^_4_dZ^  m 

dx        dr    dx  ~*~  dp    dx  "*"  a0    az 

Similarly         3F  =  a7  6r       57  a^       dF  a0 

dy        dr    dy  ~*~  d<p    dy  "*"  d0    ay 

and  ^Z  _  ?Z  ^r    ^Z  ^     ^Z  ^ 

a2    ~  a/-   a^  +  a^  dz  +  a<?   a^ 

The  second  partial  derivative  of  V  with  respect  to  x  is  obvi- 
ously obtained  from  (3)  as  follows: 

aF  av      ar  _a^  /ev\    _  dv_  av 

'  ' 


/e\    _ 

\dr)  '  ' 


dx2        dr  dx2       dx  dx\dr  d<p   dx2       dx  dx\  d 


,dVWded_/dV\ 

dO  dx2  T  dx  dx\dd/       v  ; 


319 


320 


ELECTRICAL  ENGINEERING 


dV    dV  dV 

In  equation  (6)  — ,  -r—  and  — r  are  each  functions  of  r,  <p  and  6. 
or     d(  ou 


and, 


d  fdV\ 

d2V  dr  d2V  d<p  d2V  dd 

dx  \  dr  / 
d  (dV\ 

dr2  dx  '  drdtp  dx  '  drdd  dx 
d2V  dr  d2V  dip  .d2V  dd 

dx  \  d<f>/ 

d<pdr  dx  dp2  dx  d<pdB  dx 
d2V  dr  ,  d2V  dtp  ,  d2V  dS 

(7) 


Substituting  these  values  in  equation  (6)  we  get: 

dw     dV  d*r     dV  ay     dV  d2e 

' 


dr 


_ 

~ 


^     / 
dr2  \dx 


2d2V  dr_d<p       2d2V  dO  dr 


d6  dx2 
Q2V  /dS\ 
d02\dx) 
2d2V  d0 


drd<p  dx  dx         drdO  dx  dx          d<pdO  dx 
A  similar  expression  is,  of  course,  obtained  for 

d2V        .  d2V 

—f  -n  and 


(8) 


dz2 

A  complete  discussion  of  partial  differentiation  can  be  found 
in  any  text-book  on  Calculus,  for  instance,  in  volume  II  of  WOODS 
AND  BAILEY'S  "A  Course  in  Mathematics." 

As  an  application  of  the  above  is  given  the  transformation  of 
LAPLACE'S  equation  from  rectangular  coordinates  to  spherical  and 
cylindrical  coordinates. 


FIG.  159. 

(a)  Transformation  of  LAPLACE'S  equation  to  spherical  coordi- 
nates.    Fig.  159. 


"          " 


dx2        dy2        dz2 


APPENDIX 


321 


F  =  F(rM,  r  =  fL(xyz),  B  =  fz(xyz),  <p  =  f*(xyz), 
dV        dV     dr         dV  de        dV  dtp 
~dx  ~~  ~dr     dx   "  dO  dx        dtp  dx 


dx2  =~~   dr  ax2  ~*~  dx  dx\dr 

dV  av 


" 


dx2 


But 


de  dx2  '  ax 

dx  \d<p. 


A/^Z\ 
~dx  \dO/ 


dr2  dx  "^  *~*°  *~~ 


dx  \  dip)         d<pdr  dx 


-  _L 

' 


_i        . 
ax     a0a<?  ax 


dX 


=  aF  av     aF 

'  '  dx2  '     dr  dx2  4"    de 
/dr\ 


aF 


ax2 


d_V_  /dr\         d_V_  /d6\         d_F 
ar2  \dx/  de2  \dx/  dv2 


2a2F  ar  de     2d2v  dr_  d<p 


dx, 

2d2v  de 


drde  dx  dx        drd<p  dx  dx        d6d<p  dx  dx 

Q2y  A2V 

Similar  expressions  can  be  gotten  for  -—-r  and 


'  ax2         dy 

a2F 
ar2 


dy'' 

~  ~dr  (dx2       dy2       dz2 


dz2 


dy 


aF 
"  " 


av 


-.      a2F 
"a^ 


L  5!Zr  /M  2  4.  /<M  2  •  /M  2i 

~  a<2L\ax/       "  \d)       '  \dz)  J 


dz2/ 

+  (?y 

dx  +  dy  dy 


)x  ax       dy  dy       dz  dzJ 
but  in  the  spherical  coordinate  system, 

\/x2  4-  i/2  if 

r  =  (x2  +  y2  +  22)^,  6  =  arc  tan -  and  ^  =  arc  tan  -• 


dy 

dr  30~i  ,  2a2Frar  a<^     ar  a^     ar  a^ 
""  ax  +  d  ~d      dz  a^ 


322  ELECTRICAL  ENGINEERING 

From  the  construction,  Fig.  159, 

x  =  \/x2  +  y2  cos  <p  =  r  sin  6  cos  <p',y  =  r  sin  d  sin  <f>\  z  =  r  cos  6. 

x 

2\K  =  -  =  sin  0  cos  v? 
2 


•• 
dx 


y        • 

=  ^  =  sm  0  sin 


dy       (x2  +  t/2  +  z*)*       r 
dr  _  2  _  z 

dz  ~  (x2  +  y2  +  «»)W  ~  r  = 

I^A.(X2      I      ^2)^ 

/2       ^      r  y  J 


X 


dx  x2  +  y*  x*  +  y2  +  z*  "  (x* 


Z  Z   COS  <f>  COS   ^    COS    <p 

-r  COS    V?    =     -    —         -    =     -- 

r2  r      r  r 

y 


dy  x2  +  2/2  +  22       (a;2  +  y2)* 

cos  0  sin  <p 


dz  , 


sn 


A/i\ 

y  ~dx  \xl   =  y  _         y  r 

f  xz  +  y*        ~  Vx2  +  y2      rVx2 

r 


sn 


r  sn 
1 
d<p  x  x  x  r  cos  <p 


y2       rx2  +  y2       r  sn 


a    ,  .  dO  d<p 

=  —  (sin  6  cos  <p  )  =  cos  6  cos  <p  —  —  sin  0  sin  <p  —  = 

C/^C  o3/  C/2/ 

.  cos  0  cos  <p   .     .     .    .         sin  <p 
=  cos  6  cos  v  X  -  -  +  sin  0  sin 


I          fjll.1.     \J     OX  A  A     V^  *  f*. 

r  ^  r  sin  6 


-  [cos2  0  cos2  <p  -f  sin2 


APPENDIX  323 


=  -  [(1  -  sin2  0)(1  -  sin2  <p)  +  sinV] 
=  -  [1  -  sin2  0  cos2  <f>] 

dy*  =  dy  (sin  °  sin  ^  =  =  r  [1  ~  si 

av       d  na0         i 

^=^cos0  ,       -sm0-  =    +  -sin'0 
d20        6   cos  6  cos  <p  /      1  \  dr 

>  =      -  —       =  cos  «  cos  "  cos 


1  . 

2  (cos  0  cos  <p  sm  0  cos  <p) 

i    1  /  .        cos  0  cos  <p 

H —  I  —  cos  <p  sm  0  - 
r  \  r 

.    1         ,    .         sin  <f> 
H —  cos  0  sm 


r  r  sm  0 

=  —  I  -j  sin  ^  cos  ^  cos2<^  +  sin  9  cos  0  cosV 
cos  0    .        "I 

r— — r  Sin2  <^> 

sm   0 

d   1  1  r 

=  ^-  -  cos  0  sm  $?  =   •  •  •  =  -  -I   2  sin  0  cos  0  sin2 


cos  0            i 

-   COS    <p  1 

sm  0 

a/      sin  0\ 

2    . 

92  \         r    / 

r2S1 

n     cos  V 

a  /  sin  v?  \ 

1   sin  <p  c 

>r        1  (—  sin  (p  cos  0 

r2  sin  0  fa       r  sin2  0 

cos  <     d 


r  sin  0  dx 

sin  <^g  dr       1  sin  <p  cos  0  <90       cos 


r2  sin  0  do;       r       sin2  0       dx      r  sin  0  dz 
J^  /  cos  <p  l^  cos  y?  d(p  _    1  cos  <p  cos  0  60 

~  &y  \r  sin  0  ~        r2  sin  B  dy       r        sin2  0       dy 


sn 


r  sin  0 


21 


324  ELECTRICAL  ENGINEERING 


1   sin  (p    .  1  sin  <p  cos  0 

^  STl  sin  '  cos  *  +  F     sin*  a 


r 

r  sin  0  r  sin  0       r2  sin  0 

i   a  sm   (p 

1  cos  <p  cos  0  cos  0  sin  9 

•>        sin  <p    cos  <p 

r        sin2  0                r 

r  sin  0  r  sin  6 

av     a2 

'  *  az2  +  di 

r     av  _ 

1  2 

-[1  —  sin2  0  cos2  (p  +  1  —  sin2  <^  +  sin2  0]  =  - 


cos  0 


ax2       dy2       dz2  r2      sin  0 

W    +  \a?//      "  \dz> 

sin2  0  cos2  </>  +  sin2  0  sin2  <p  +  cos2  0  =  1 

/a0\  2     ,d0\  2     /a0\  2  =  i 

W         \dyl         \dz'  r2  sin2  0 

/a<p\  2     /av?\  2     /a<p\  2  _  i 

\a^/      "  (dy)      "  Vaz/  ~  r2  sin2  0 

ar  a0  ,  ar  a0  ,  ar  a0     ix 

T-^-  +  T-  T,-  +  T-^  =  -(sin  0  cos  0  cos2  <p  -f-  sm  0  cos  0  sm2v?  — 
ox  ox      oil  dy      dz  oz      r  .  . 

sin  0  cos  0)  =  0 

dx  dx       dy  dy       dz  dz 

d0  d(p       d0  dtp       dd  d<p 


a2F    a2F    a2F  =  aFrav    av    a2ri    aFra20    a20    a20 
~  ~" 


an  2    /an  2    /an  2-|    a2Fr  /a0\  2    /a0\  2    /a0\ 
dx)      (dy)   hW  J^a^Lvax/      w      \dJ 

v,  JFr  /a^\  2    /a<^\  2    /a<p\  2"i  _2  aF  i  cos  0  aF 

^X^l  \^/   '•  \dv/        \dz)  .  ~r  ~dr  r2  sin  0  a0 


.  i_  a2F 

-        + 


sin2  0 


i  r  /0aF  ,    a2F\  .    i    /     flaF, 
Ar  (2a7+r  ai^)  +sirT0  (cos  g  a0-+s 

A  B 


APPENDIX 


325 


But 


.'.  LAPLACE'S  equation  in  spherical  coordinates 


sin2  $  d<p* 


(C)  CYLINDRICAL  COORDINATES 

Referring  in  Fig.  160 
V  =  F(rBz)    r  =  (z2  +  y*)Kf    e  =  tan'  ^z  =  z 


x  =  r  cos  e,  y  =  r  sin  0  z  =  z 

—  =  x  -  x 

dx  ~  (x2  -f  y2)^  ~  r 

dr 
dy 

dr 

dz 


cos  6 

=  sin  6 

=  0 


< 

Six* 


FIG.  160. 


60 
dx 


+© 


y 

r~ 


sn 


cos 


326  ELECTRICAL  ENGINEERING 


?-« 

dz 

dz_dz__fa_^ 
dx        dy  dz 

av      a  ,a0         sin202 

—  =  —  cos  0  =  —  sm  0  —  =  H — 
ax2       az  dx  r 

av  cos2  0 


^-r-o 

«        O       *-* 

az2 

a20        a        sin  0  1         n  d8    .    1         n  ar        2 

v-s  = = sm  0  —  +  —  sm  0  -r-  =  —  sm  0  cos  0 

aar       aa;          r  r  ao;       r^  dx       r2 

a20      a  cos  0         i        so     i         a0         2 

^-s  =  —         -  = sm  0 cos  0  --  = 5  sm  0  cos  0 

a?/z       dy      r  r  dy       r2-  dy  r2 

^  =  0 

az2 

a *z  =  d*z  =  d%  = 

dx2       dy2       dx2 

Vr     av     av  = •  i 

'  '  dx2       dy2       dz2  ~  r 

a20     a20     a20 

dx2       8v*       8z2~ 


(I)'  +©'+©•- 

dr<Wd^rdOdrd<)_ 
dx  dx       dy  dy       dz  dz 
dr  dz        dr  dz       dr  dz 

+  " 


= 

dx  dx  +  a?/  dy       dz  dz  ~ 


'  a^2      a?/2      a^2  ~  r  ar2      ar2      r2  a02  ""  a^2  = 

which  is  LAPLACE'S  equation  in  cylindrical  coordinates. 


APPENDIX  II 

Elements  of  Vector  Analysis. — Physical  quantities  can  be 
divided  into  two  large  and  important  classes,  namely:  scalars 
and  vectors. 

A  scalar  quantity  is  one  that  is  absolutely  determined  by  its 
magnitude.  Thus  temperature,  work,  etc.,  are  scalars. 

A  vector  quantity  may  be  denned  as  one  having  magnitude, 
sense  and  direction  and  it  is  necessary  to  specify  these  three  in 
order  to  determine  a  vector.  Velocities  and  accelerations  are 
examples  of  vector  quantities;  forces  are  strictly  not  vectors, 
since  they  are  characterized  not  only  by  their  magnitude,  sense 
and  direction  but  also  by  the  point  of  application,  while  vectors 
do  not  have  definite  position  in  space.  However,  forces  can 
be  treated  as  vectors  when  proper  account  is  taken  of  this 
difference. 

Addition  and  Subtraction  of  Vectors. — Vectors  are  added  or 
subtracted  by  the  well-known  parallelogram  law: 

Thus 

a  +  o  =  c 

and 

c  —  o  —  a. 

Vectors  follow  the  associative  and  com- 
mutative laws  of  algebra,  and  hence  very 
little  explanation  is  necessary  as  to  the 
addition  of  vectors.  a 

The  sum  of  three  vectors  a,  b  and  c  is  FlG>  161> 

given   by  the   diagonal  mn  as  shown  in  Fig.  161. 

Products  of  Vectors. — There  are  two  kinds  of  vector  products : 

I.  The  dot  product  which  is  denned  as, 

a  dot  b  =  a  •  b  =  ab  cos  (a,  b) 

where  a  and  b  are  the  two  vectors  to  be  multiplied  together, 
and  a  and  b  are  the  numerical  values  of  the  vectors. 

II.  The  cross  product  which  is  denned  as: 

a  cross  b  =  a  X  b  =  e  ab  sin  (a,  b) . 
327 


328  ELECTRICAL  ENGINEERING 

where  e  denotes  that  the  product  is  a  vector.  It  is  the  unit 
vector  perpendicular  to  the  plane  formed  by  a  and  6. 

The  above  names  have  been  introduced  by  WILLABD  GIBBS 
and  they  are  used  principally  by  American  writers. 

The  reader  is  familiar  with  the  resolution  of  vectors  into  com- 
ponents which  can  be  treated  according  to  the  laws  of  ordinary 
algebra.  The  great  advantage  of  vector  analysis  is  that  it  deals 
with  vectors  directly.  It  is  found  useful,  however,  to  resolve 
vectors  into  their  components  and  in  such  case  a  vector  a  is 
defined  in  terms  of  its  magnitude  along  any  direction,  say  x, 
times  a  unit  vector  i  along  x. 

For  convenience  rectangular  coordinates  are  used  and  the 
unit  vector  along  the  z-axis  is  denoted  by  i,  the  unit  vector  along 
the  y-axis  is  denoted  by  j  and  the  unit  vector  along  the  z-axis 
byfc. 

Thus 

a  =  axi  -f-  avj  -f-  axk 

a  =  A/a*2  +  av2  +  a22 

a  =  a(i  cos  a  -f-  j  cos  /3  -f  k  cos  7) 
where  a,  £  and  7  are  the  direction  cosines. 

Now  it  will  be  easily  seen  from  the  definition  of  the  dot  product 
that: 

i  -  i  —  1  i  -  j  =  0 

j-j  =  l  t-fc  =  0 

k-k  =  1  j  -k  =  0 

a  -  a  =  a2 

It  is  also  clear  that  the  condition  of  perpendicularity  of  two 
vectors  is  that  their  dot  product  shall  be  zero. 
The  dot  product  is  also  called  (by  HAMILTON) 
the  scalar  product,   because  the   product  is  a 
scalar.     The  cross  product  is  called  the  vector 
product,  because  it  is  a  vector. 
.a      a  X  6  gives  a  vector  c,  Fig.  162,  whose  mag- 
nitude is  (ab)  sin  (a,6);   its  direction  is  along 
the  normal  to  the  plane  of  the  vectors  a  and  6, 
and  finally  the  sense  of  c  is  taken  so  that  as  one 
goes  from  a  to  6  he  follows  a  right-hand  screw.     In  other  words 
from  a  to  b  we  follow  the  threads  of  a  corkscrew  whose  direction 
of  progress  determines  the  sense  of  6.    This  is,  of  course,  the  well- 


APPENDIX 


329 


known  rule  of  MAXWELL  for  the  relation  between  the  direction 
of  flux,  the  motion  of  a  conductor,  and  the  e.m.f.  thereby 
generated. 

It  is  clear  from  the  definition  of  a  cross  product  that  in  Fig.  163 

i  X  j  =  k  =  -  j  Xi 
j  x  k  =  i  =  -  k  X  j 
k  X  i  =  j  =  --  i  X  k 
i  X  *  =  0  j  X  j  =  0  k  X  k  =  0. 


FIG.  163. 

The  cross  product  of  two  vectors  can  also  be  obtained  in 
terms  of  the  components  and  the  unit  vectors  i,  j  and  /b;  only 
it  is  evident  that  care  should  be  taken  not  to  invert  the  order 
of  factors,  since  a  Xb  =  —  6  X  a. 

Exercise. — Prove  that  if  ax  av  a2,  bx  bv  bz  are  the  rectangular 
components  of  a  and  b. 

a  X  b  =  (avbz  -  aj)v)  i  +  (afbx  -  ajb,)  j  +  (axbv  -  avbx)k 
or  in  determinant  form, 

*  j 

a  X  b  =  axav 
bx  by  bz 

Exercise. — Prove  that  the  absolute  value  of  a  X  b  which  is 
written  a  X  b 

=  (a)  (b)  sin  (a,  b) 


a  X  6= 


bv2  +  bf2)  -  (axbx  +  avb 


330  ELECTRICAL  ENGINEERING 

Now  it  will  be  noticed  that  in  the  last  exercise,  a2x  +  a2y  +  az2 
is  simply  equal  to  a  •  a. 

Thus: 

First  term  =  a  •  a 

Second  term          =  b  -  b 

Third  term  =  (a  •  6)2 

=  (ab  cos  a)2 
where  =  <£  a,  b 

so  that  a  X  &   =  V(a  •«)(&•  &)  -  (ab  cos  a)2 


a  X  6  =  "a2&2  -  a262  cos2  a 


=  a&   \  1  —  cos  a 
=  ab  sin  a 

The  product  of  a  X  6,  must  be  the  normal  to  the  plane  of  the 
vectors  a  and  b  is  seen  as  follows  :  Assume  c  to  be  the  vector  and 
find  a  •  c  =  a.  (a  X  b) 

also  b-c  =  b.     (a  X  b) 

Multiplying  these  out  in  the  ordinary  way  we  find 

a  •  c  =  0     6  •  c  =  0, 
i.e.,         ac  cos  (a,  c)  =  0 
be  cos  (bj  c)  =  0 

which  is  satisfied  when  c  is  normal  to  the  plane  ab. 

The  above  are  intended  to  cover  the  very  small  part  of  vector 
analysis  used  in  this  book.  For  further  information  the  reader 
should  consult  special  treatises  written  on  the  subject. 

HEAVISIDES'  "  Electromagnetic  Theory;"  ABRAHAM  and 
FOPPL'S  "Theory  of  Electricity  and  Magnetism"  can  be  recom- 
mended highly. 

An  excellent  short  treatise  on  the  subject  is  "Elements  of 
Vector  Analysis"  by  BURALLI-FORTI  and  R.  MAREOLONGO,  and 
a  somewhat  larger  work  is  that  of  WILLARD  GIBBS,  edited  by 
WILSON.  Finally  COFFIN'S  "Vector  Analysis"  may  be  men- 
tioned among  works  of  reference,  it  appears  indeed  as  best  suited 
for  the  introduction  to  vector  analysis. 


INDEX 


Attenuation,  121 

Austin,  316 

Auxiliary  equations,  46 

B 

Ber  and  bei  function,  273 
Bjerknes,  313 


Capacity    between    concentric    con- 
ductors, 70 

between  parallel  planes,  68 

between  transmission  lines,  70 

of   antenna,  299 

of  a  single  wire,  230 

of  concentric  cable,  96 

of  isolated  spheres,  163 

of  two   cylindrical  conductors, 
224 

of  two  wires  in  multiple,  230 
Charge  distribution  on  an  ellipsoid, 

199 

Circular  symmetry,  188 
Complete  differential,  164 
Complimentary  function,  46,  86 
Concentric  cylinders,  215 

spheres,  209 
Condenser,  capacity  of,  68 

characteristics  of,  68 

charged,  71 

discharged,  72 

energy  supplied  to,  71 
Coulomb's  law,  157 
Coupling,  effect  on  frequency,  303 
Curl  of  a  vector,  257 
Current,  equation  of,  261 
Curvature  of  earth,  314 


Cylindrical  bars,  150 

conductor,     current     and     flux 

distribution,  268 
conductors,  218 

D 

Differential  equations,  higher  order, 
44 

operator,  D,  45 

Differentials  and  differences,  61 
Direct-current   generator,    field    cir- 
cuit, 59 

Displacement  current,  263,  264 
Distortionless  line,  142 
Divergence  of  a  vector,  185 

theorem,  186 


E 


Electric  doublet,  284 

field,  energy  of,  262 

intensity,  290 

Electromagnetic  radiation,  278 
Electromotive  force,  equation  of,  260 

F 

Field  intensity,  158 

Flat  conductors,  152  . 

Flux  and  current  distribution,  152 

Forces  between  point  charges  and 

spheres,  175 
Froelich's  equation,  22 
Fuller,  316 


G 


Gauss'  theorem,  160 
Graph  of  function  y  =  <rx,  9 
Green's  theorem,  186 
Grounded  horizontal  wire,  effect  on 
potential  distribution,  254 


331 


332 


INDEX 


H 

Hertz's  oscillator,  284 
Hysteresis  loop,  56 

I 

Images,  method  of,  168 
Inductance,  11 

of  air  coil,  36 

of  concentric  cables,  97 
Inductances,  combined,  37 
Inductive  circuit  containing  iron,  62 


K 


Kelvin,  273 


Lame's  differential  parameter,  185 
Laplace's  equation,  188,  320 
Legendre's  coefficient,  188 

function,  189 
Leyden  jars,  72 
Linear  differential  equations,  3 
Line  charge,  218 

integral,  163 
Logarithmic  decrement,  295 

M 

Magnetic  field,  energy  of,  263 

energy  stored  in,  8 

intensity,  290 

potential,  180 

and  current,  183,  184 

shell,  181 
Marconi,  292 

Maxwell's  coefficient,  232,  263 
Metallic  spheres,  169 
Mutual  induction,  33 

imperfect,  51 

perfect,  37 

O 

Oblate  ellipsoid,  potential  distribu- 
tion, 24 


Partial  differentiation,  319 
fractions,  21 


Poisson's  equation,  187 
Potential,  162 

distribution       between       point 

charge  and  plane,  171 
between  two  spheres,  175 

gradient,  164,  165 

of  small  magnet,  180 

outside  of  thin  circular  disc,  197 
Power  factor,  295 

received,  317 

R 

Radiated  energy,  290 
Radiation,  278 

resistance,  292 
Receiving  station,  313 

S 

Short-circuited  winding,  current  in, 

54 

Short-circuit  suddenly  opened,  84 
Shunt  motor  self  excited,  16 
Skin  effect,  271 
Solenoidal  field,  186 
Solid  angle,  183 
Step-by-step  method,  66 
Stoke's  theorem,  258 
Surface  density,  170 

integral  of   distributed   vector, 

158 
Symbolic  factors,  89 


Three-phase  cable,  243 
Three-phase  line,  249 
Tuned  circuit,  80 
Two  conductor  cables,  237 

U 

Unit  charge  and  unit  pole,  157 
V 

Vector  analysis,  327 

Velocity  of  propagation,  121-132 

W 

Wave  lengths,  121-132 
Weber's  equation,  182 


THIS  BOOK  IS  DUE  ON  THE  LAST  BATE 
BELOW 


AN     INITIAL    FINE      OI 

WILL.  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  «1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


OCT   23   1932 


CT 
WOV 


8    1933 
3CT    5  1934 

3   194! 
L^St- 


dc 

INTER-L  3RAR1 


a 

f  LOAN 


LD  21-50m-8,-32 


YC   19681 


84283 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


